# Two-Sample Inference

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Chapter 9: Two-Sample Inference Chapter 7 discussed methods of hypothesis testing about one-population parameters. Chapter 8 discussed methods of estimating population parameters from one sample using confidence intervals. This chapter will look at methods of confidence intervals and hypothesis testing for two populations. Since there are two populations, there are two random variables, two means or proportions, and two samples (though with paired samples you usually consider there to be one sample with pairs collected). Examples of where you would do this are:

Testing and estimating the difference in testosterone levels of men before and after they had children (Gettler, McDade, Feranil & Kuzawa, 2011).

Testing the claim that a diet works by looking at the weight before and after subjects are on the diet. Estimating the difference in proportion of those who approve of President Obama in the age group 18 to 26 year olds and the 55 and over age group.

All of these are examples of hypothesis tests or confidence intervals for two populations. The methods to conduct these hypothesis tests and confidence intervals will be explored in this method. As a reminder, all hypothesis tests are the same process. The only thing that changes is the formula that you use. Confidence intervals are also the same process, except that the formula is different. Section 9.1 Two Proportions There are times you want to test a claim about two population proportions or construct a confidence interval estimate of the difference between two population proportions. As with all other hypothesis tests and confidence intervals, the process is the same though the formulas and assumptions are different. Hypothesis Test for Two Population Proportion (2-Prop Test)

1. State the random variables and the parameters in words. x1 = number of successes from group 1 x2 = number of successes from group 2 p1 = proportion of successes in group 1 p2 = proportion of successes in group 2

2. State the null and alternative hypotheses and the level of significance

Ho : p1 = p2 or Ho : p1 − p2 = 0 HA : p1 < p2 HA : p1 > p2 HA : p1 ≠ p2

HA : p1 − p2 < 0 HA : p1 − p2 > 0 HA : p1 − p2 ≠ 0

Also, state your α level here.

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3. State and check the assumptions for a hypothesis test

a. A simple random sample of size n1 is taken from population 1, and a simple random sample of size n2 is taken from population 2.

b. The samples are independent. c. The assumptions for the binomial distribution are satisfied for both

populations. d. To determine the sampling distribution of p̂1 , you need to show that n1p1 ≥ 5

and n1q1 ≥ 5 , where q1 = 1− p1 . If this requirement is true, then the sampling distribution of p̂1 is well approximated by a normal curve. To determine the sampling distribution of p̂2 , you need to show that n2p2 ≥ 5 and n2q2 ≥ 5 , where q2 = 1− p2 . If this requirement is true, then the sampling distribution of

p̂2 is well approximated by a normal curve. However, you do not know p1 and p2 , so you need to use p̂1 and p̂2 instead. This is not perfect, but it is the best you can do. So you just need to make sure that x1 , n1 − x1 , x2 ,and n2 − x2 are all more than 5.

4. Find the sample statistics, test statistic, and p-value

Sample Proportion: n1 = size of sample 1

p̂1 = x1 n1

(sample 1 proportion)

q̂1 = 1− p̂1 (complement of p̂1)

n2 = size of sample 2

p̂2 = x2 n2

(sample 2 proportion)

q̂2 = 1− p̂2 (complement of p̂2 )

Pooled Sample Proportion, p :

p = x1 + x2 n1 + n2

q = 1− p

Test Statistic:

z = p̂1 − p̂2( )− p1 − p2( )

pq n1

+ pq n2

Usually p1 − p2 = 0 , since Ho : p1 = p2

p-value: Use normalcdf(lower limit, upper limit, 0, 1) (Note: if HA : p1 < p2 , then lower limit is −1E99 and upper limit is your test statistic. If HA : p1 > p2 , then lower limit is your test statistic and the upper limit is 1E99 . If HA : p1 ≠ p2 , then find the p-value for HA : p1 < p2 , and multiply by 2.)

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5. Conclusion This is where you write reject Ho or fail to reject Ho . The rule is: if the p-value < α , then reject Ho . If the p-value ≥α , then fail to reject Ho

6. Interpretation

This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show HA is true, or you do not have enough evidence to show HA is true.

Confidence Interval for the Difference Between Two Population Proportion (2-Prop Interval) The confidence interval for the difference in proportions has the same random variables and proportions and the same assumptions as the hypothesis test for two proportions. If you have already completed the hypothesis test, then you do not need to state them again. If you haven’t completed the hypothesis test, then state the random variables and proportions and state and check the assumptions before completing the confidence interval step.

1. Find the sample statistics and the confidence interval Sample Proportion:

n1 = size of sample 1

p̂1 = x1 n1

(sample 1 proportion)

q̂1 = 1− p̂1 (complement of p̂1)

n2 = size of sample 2

p̂2 = x2 n2

(sample 2 proportion)

q̂2 = 1− p̂2 (complement of p̂2 )

Confidence Interval:

The confidence interval estimate of the difference p1 − p2 is p̂1 − p̂2( )− E < p1 − p2 < p̂1 − p̂2( ) + E

where the margin of error E is given by E = zc p̂1q̂1 n1

+ p̂2q̂2 n2

zC = critical value where C = 1−α

2. Statistical Interpretation: In general this looks like, “there is a C% chance that p̂1 − p̂2( )− E < p1 − p2 < p̂1 − p̂2( ) + E contains the true difference in proportions.”

3. Real World Interpretation: This is where you state how much more (or less) the

first proportion is from the second proportion. The critical value is a value from the normal distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the sample proportion, and the interval has a probability of being true, then you can think of this as the statement P p̂1 − p̂2( )− E < p1 − p2 < p̂1 − p̂2( ) + E( ) = C . So you can use the invNorm command on

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the calculator to find the critical value. These are always the same value, so it is easier to just look at the table A.1 in the Appendix. Example #9.1.1: Hypothesis Test for Two Population Proportions

Do husbands cheat on their wives more than wives cheat on their husbands (“Statistics brain,” 2013)? Suppose you take a group of 1000 randomly selected husbands and find that 231 had cheated on their wives. Suppose in a group of 1200 randomly selected wives, 176 cheated on their husbands. Do the data show that the proportion of husbands who cheat on their wives are more than the proportion of wives who cheat on their husbands. Test at the 5% level. Solution:

1. State the random variables and the parameters in words. x1 = number of husbands who cheat on his wife x2 = number of wives who cheat on her husband p1 = proportion of husbands who cheat on his wife p2 = proportion of wives who cheat on her husband

2. State the null and alternative hypotheses and the level of significance

Ho : p1 = p2 or Ho : p1 − p2 = 0 HA : p1 > p2 HA : p1 − p2 > 0 α = 0.05

3. State and check the assumptions for a hypothesis test

a. A simple random sample of 1000 responses about cheating from husbands is taken as stated. A simple random sample of 1200 responses about cheating from wives is taken as stated.

b. The samples are independent since different genders. c. The assumptions for the binomial distribution are satisfied in both

populations, since there are only two responses, there are a fixed number of trials, the probability of a success is the same, and the trials are independent.

d. x1 = 231, n1 − x1 = 1000 − 231= 769 , x2 = 176 , and n2 − x2 = 1200 −176 = 1024 are all greater than or equal to 5. So both sampling distributions of p̂1 and p̂2 can be approximated with a normal distribution.

4. Find the sample statistics, test statistic, and p-value Sample Proportion:

n1 = 1000

p̂1 = 231 1000

= 0.231

q̂1 = 1− 231 1000

= 769 1000

= 0.769

n2 = 1200

p̂2 = 176 1200

≈ 0.1467

q̂2 = 1− 176 1200

= 1024 1200

≈ 0.8533

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Pooled Sample Proportion, p :

p = 231+176 1000 +1200

= 407 2200

= 0.185

q = 1− 407 2200

= 1793 2200

= 0.815

Test Statistic:

z = 0.231− 0.1467( )− 0 0.185*0.815

1000 + 0.185*0.815

1200 = 5.0704

p-value: normalcdf 5.0704,1E99,0,1( ) = 1.988 ×10−7

Figure #9.1.1: Setup for 2-PropZTest on TI-83/84 Calculator

Figure #9.1.2: Results for 2-PropZTest on TI-83/84 Calculator

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5. Conclusion Reject Ho , since the p-value is less than 5%.

6. Interpretation

This is enough evidence to show that the proportion of husbands having affairs is more than the proportion of wives having affairs.

Example #9.1.2: Confidence Interval for Two Population Proportions

Do more husbands cheat on their wives more than wives cheat on the husbands (“Statistics brain,” 2013)? Suppose you take a group of 1000 randomly selected husbands and find that 231 had cheated on their wives. Suppose in a group of 1200 randomly selected wives, 176 cheated on their husbands. Estimate the difference in the proportion of husbands and wives who cheat on their spouses using a 95% confidence level. Solution:

1. State the random variables and the parameters in words. These were stated in example #9.3.1, but are reproduced here for reference. x1 = number of husbands who cheat on his wife x2 = number of wives who cheat on her husband p1 = proportion of husbands who cheat on his wife p2 = proportion of wives who cheat on her husband

2. State and check the assumptions for the confidence interval

The assumptions were stated and checked in example #9.3.1.

3. Find the sample statistics and the confidence interval Sample Proportion:

n1 = 1000

p̂1 = 231 1000

= 0.231

q̂1 = 1− 231 1000

= 769 1000

= 0.769

n2 = 1200

p̂2 = 176 1200

≈ 0.1467

q̂2 = 1− 176 1200

= 1024 1200

≈ 0.8533

Confidence Interval:

zC = 1.96

E = 1.96 0.231*0.769 1000

+ 0.1467*0.8533 1200

= 0.033

The confidence interval estimate of the difference p1 − p2 is p̂1 − p̂2( )− E < p1 − p2 < p̂1 − p̂2( ) + E 0.231− 0.1467( )− 0.033< p1 − p2 < 0.231− 0.1467( ) + 0.033 0.0513< p1 − p2 < 0.1173

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Figure #9.1.3: Setup for 2-PropZInt on TI-83/84 Calculator

Figure #9.1.4: Results for 2-PropZInt on TI-83/84 Calculator

4. Statistical Interpretation: There is a 95% chance that 0.0513< p1 − p2 < 0.1173 contains the true difference in proportions.

5. Real World Interpretation: The proportion of husbands who cheat is anywhere

from 5.13% to 11.73% higher than the proportion of wives who cheat. Section 9.1: Homework In each problem show all steps of the hypothesis test or confidence interval. If some of the assumptions are not met, note that the results of the test or interval may not be correct and then continue the process of the hypothesis test or confidence interval. 1.) Many high school students take the AP tests in different subject areas. In 2007, of

the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female (“AP exam scores,” 2013). Is there enough evidence to show that the proportion of female students taking the biology exam is higher than the proportion of female students taking the calculus AB exam? Test at the 5% level.

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2.) Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female (“AP exam scores,” 2013). Estimate the difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam using a 90% confidence level.

3.) Many high school students take the AP tests in different subject areas. In 2007, of the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male (“AP exam scores,” 2013). Is there enough evidence to show that the proportion of female students taking the calculus AB exam is different from the proportion of male students taking the calculus AB exam? Test at the 5% level.

4.) Many high school students take the AP tests in different subject areas. In 2007, of

the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male (“AP exam scores,” 2013). Estimate using a 90% level the difference in proportion of female students taking the calculus AB exam versus male students taking the calculus AB exam.

5.) Are there more children diagnosed with Autism Spectrum Disorder (ASD) in

states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated (“Autism and developmental,” 2008). Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah? Test at the 1% level.

6.) Are there more children diagnosed with Autism Spectrum Disorder (ASD) in

states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated (“Autism and developmental,” 2008). Estimate the difference in proportion of children diagnosed with ASD between Pennsylvania and Utah. Use a 98% confidence level.

7.) A child dying from an accidental poisoning is a terrible incident. Is it more likely

that a male child will get into poison than a female child? To find this out, data was collected that showed that out of 1830 children between the ages one and four who pass away from poisoning, 1031 were males and 799 were females (Flanagan, Rooney & Griffiths, 2005). Do the data show that there are more male children dying of poisoning than female children? Test at the 1% level.

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8.) A child dying from an accidental poisoning is a terrible incident. Is it more likely that a male child will get into poison than a female child? To find this out, data was collected that showed that out of 1830 children between the ages one and four who pass away from poisoning, 1031 were males and 799 were females (Flanagan, Rooney & Griffiths, 2005). Compute a 99% confidence interval for the difference in proportions of poisoning deaths of male and female children ages one to four.

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Section 9.2 Paired Samples for Two Means Are two populations the same? Is the average height of men taller than the average height of women? Is the mean weight less after a diet than before? You can compare populations by comparing their means. You take a sample from each population and compare the statistics. Anytime you compare two populations you need to know if the samples are independent or dependent. The formulas you use are different for different types of samples. If how you choose one sample has no effect on the way you choose the other sample, the two samples are independent. The way to think about it is that in independent samples, the individuals from one sample are overall different from the individuals from the other sample. This will mean that sample one has no affect on sample two. The sample values from one sample are not related or paired with values from the other sample. If you choose the samples so that a measurement in one sample is paired with a measurement from the other sample, the samples are dependent or matched or paired. (Often a before and after situation.) The way to think about it is that in dependent samples, the individuals from one sample are the same individuals from the other sample. This makes the sample values from each sample paired. Example #9.2.1: Independent or Dependent Samples

Determine if the following are dependent or independent samples. a.) Randomly choose 5 men and 6 women and compare their heights Solution:

Independent, since there is no reason that one value belongs to another. The individuals are not the same for both samples. The individuals are definitely different. A way to think about this is that the knowledge that a man is chosen in one sample does not give any information about any of the woman chosen in the other sample.

b.) Choose 10 men and weigh them. Give them a new wonder diet drug and later weigh them again.

Solution: Dependent, since each person’s before weight can be matched with their after weight. The individuals are the same for both samples. A way to think about this is that the knowledge that a person weighs 400 pounds at the beginning will tell you something about their weight after the diet drug.

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c.) Take 10 people and measure the strength of their dominant arm and their non- dominant arm.

Solution: Dependent, since you can match the two arm strengths. The individuals are the same for both samples. So the knowledge of one person’s dominant arm strength will tell you something about the strength of their non-dominant arm.

To analyze data when there are matched or paired samples, called dependent samples, you conduct a paired t-test. Since the samples are matched, you can find the difference between the values of the two random variables. Hypothesis Test for Two Sample Paired t-Test

1. State the random variables and the parameters in words. x1 = random variable 1 x2 = random variable 2 µ1 = mean of random variable 1 µ2 = mean of random variable 2

2. State the null and alternative hypotheses and the level of significance The usual hypotheses would be Ho :µ1 = µ2 or Ho :µ1 − µ2 = 0 HA :µ1 < µ2 HA :µ1 > µ2 HA :µ1 ≠ µ2

HA :µ1 − µ2 < 0 HA :µ1 − µ2 > 0 HA :µ1 − µ2 ≠ 0

However, since you are finding the differences, then you can actually think of µ1 − µ2 = µd , µd = population mean value of the differences So the hypotheses become Ho :µd = 0 H1 :µd < 0 HA :µd > 0 HA :µd ≠ 0 Also, state your α level here.

3. State and check the assumptions for the hypothesis test a. A random sample of n pairs is taken. b. The population of the difference between random variables is normally

distributed. In this case the population you are interested in has to do with the differences that you find. It does not matter if each random variable is normally distributed. It is only important if the differences you find are normally distributed. Just as before, the t-test is fairly robust to the assumption if the sample size is large. This means that if this assumption isn’t

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met, but your sample size is quite large (over 30), then the results of the t-test are valid.

4. Find the sample statistic, test statistic, and p-value

Sample Statistic: Difference: d = x1 − x2 for each pair

Sample mean of the differences: d = d∑ n

Standard deviation of the differences: sd = d − d( )2∑ n −1

Number of pairs: n Test Statistic:

t = d − µd

sd n with degrees of freedom = df = n −1 Note: µd = 0 in most cases.

p-value: Use tcdf lower limit, upper limit, df( ) (Note: if HA :µd < 0 , then lower limit is −1E99 and upper limit is your test statistic. If HA :µd > 0 , then lower limit is your test statistic and the upper limit is 1E99 . If HA :µd ≠ 0 , then find the p-value for HA :µd < 0 , and multiply by 2.)

5. Conclusion

This is where you write reject Ho or fail to reject Ho . The rule is: if the p-value < α , then reject Ho . If the p-value ≥α , then fail to reject Ho

6. Interpretation

This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show HA is true, or you do not have enough evidence to show HA is true.

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Confidence Interval for Difference in Means from Paired Samples (t- Interval) The confidence interval for the difference in means has the same random variables and means and the same assumptions as the hypothesis test for two paired samples. If you have already completed the hypothesis test, then you do not need to state them again. If you haven’t completed the hypothesis test, then state the random variables and means, and state and check the assumptions before completing the confidence interval step.

1. Find the sample statistic and confidence interval Sample Statistic:

Difference: d = x1 − x2

Sample mean of the differences: d = d∑ n

Standard deviation of the differences: sd = d − d( )2∑ n −1

Number of pairs: n Confidence Interval: The confidence interval estimate of the difference µd = µ1 − µ2 is

d − E < µd < d + E

E = tc sd n

tc is the critical value where C = 1−α and degrees of freedom df = n −1

2. Statistical Interpretation: In general this looks like, “there is a C% chance that the

statement d − E < µd < d + E contains the true mean difference.” 3. Real World Interpretation: This is where you state what interval contains the true

mean difference. The critical value is a value from the Student’s t-distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the sample mean, and the interval has a probability of containing the true mean difference, then you can think of this as the statement P d − E < µd < d + E( ) = C . To find the critical value, you use table A.2 in the Appendix. How to check the assumptions of t-test and confidence interval: In order for the t-test or confidence interval to be valid, the assumptions of the test must be met. So whenever you run a t-test or confidence interval, you must make sure the assumptions are met. So you need to check them. Here is how you do this:

1 For the assumption that the sample is a random sample, describe how you took the samples. Make sure your sampling technique is random and that the samples were dependent.

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2 For the assumption that the population of the differences is normal, remember the process of assessing normality from chapter 6.

Example #9.2.2: Hypothesis Test for Paired Samples Using the Formula

A researcher wants to see if a weight loss program is effective. She measures the weight of 6 randomly selected women before and after the weight loss program (see table #9.2.1). Is there evidence that the weight loss program is effective? Test at the 5% level. Table #9.2.1: Data of Before and After Weights

Person 1 2 3 4 5 6 Weight before

165 172 181 185 168 175

Weight after

143 151 156 161 152 154

Solution:

1. State the random variables and the parameters in words. x1 = weight of a woman after the weight loss program x2 = weight of a woman before the weight loss program µ1 = mean weight of a woman after the weight loss program µ2 = mean weight of a woman before the weight loss program

2. State the null and alternative hypotheses and the level of significance

Ho :µd = 0 HA :µd < 0 α = 0.05

3. State and check the assumptions for the hypothesis test a. A random sample of 6 pairs of weights before and after was taken. This was

stated in the problem, since the women were chosen randomly. b. The population of the difference in after and before weights is normally

distributed. To see if this is true, look at the histogram, number of outliers, and the normal probability plot. (If you wish, you can look at the normal probability plot first. If it doesn’t look linear, then you may want to look at the histogram and number of outliers at this point.)

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Figure #9.2.1: Histogram of Differences in Weights

This histogram looks somewhat normal. There is only one outlier in the difference data set. Figure #9.2.2: Normal Probability Plot of Differences in Weights

The probability plot on the differences looks somewhat linear. So you can assume that the distribution of the difference in weights is normal. 4. Find the sample statistic, test statistic, and p-value Sample Statistics:

Table #9.2.2: Differences Between Before and After Weights

Person 1 2 3 4 5 6 Weight after, x1

143 151 156 161 152 154

Weight before, x2

165 172 181 185 168 175

d = x1 − x2 −22 −21 −25 −24 −16 −21 The mean and standard deviation are

d = −21.5 sd = 3.15

Test Statistic:

t = d − µd sd n

= −21.5 − 0 3.15 6

= −16.779

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p-value: There are six pairs so the degrees of freedom are df = n −1= 6 −1= 5 Since H1 :µd < 0 , then p-value = tcdf −1E99,−16.779,5( ) ≈ 6.87 ×10−6

5. Conclusion Since the p-value < 0.05, reject Ho .

6. Interpretation There is enough evidence to show that the weight loss program is effective.

Note: Just because the hypothesis test says the program is effective doesn’t mean you should go out and use it right away. The program has statistical significance, but that doesn’t mean it has practical significance. You need to see how much weight a person loses, and you need to look at how safe it is, how expensive, does it work in the long term, and other type questions. Remember to look at the practical significance in all situations. In this case, the average weight loss was 21.5 pounds, which is very practically significant. Do remember to look at the safety and expense of the drug also. Example #9.2.3: Hypothesis Test for Paired Samples Using the Calculator

The New Zealand Air Force purchased a batch of flight helmets. They then found out that the helmets didn’t fit. In order to make sure that they order the correct size helmets, they measured the head size of recruits. To save money, they wanted to use cardboard calipers, but were not sure if they will be accurate enough. So they took 18 recruits and measured their heads with the cardboard calipers and also with metal calipers. The data in centimeters (cm) is in table #9.2.3 (“NZ helmet size,” 2013). Do the data provide enough evidence to show that there is a difference in measurements between the cardboard and metal calipers? Use a 5% level of significance.

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Table #9.2.3: Data for Head Measurements Cardboard Metal

146 145 151 153 163 161 152 151 151 145 151 150 149 150 166 163 149 147 155 154 155 150 156 156 162 161 150 152 156 154 158 154 149 147 163 160

Solution:

1. State the random variables and the parameters in words. x1 = head measurement of recruit using cardboard caliper x2 = head measurement of recruit using metal caliper µ1 = mean head measurement of recruit using cardboard caliper µ2 = mean head measurement of recruit using metal caliper

2. State the null and alternative hypotheses and the level of significance

Ho :µd = 0 HA :µd ≠ 0 α = 0.05

3. State and check the assumptions for the hypothesis test a. A random sample of 18 pairs of head measures of recruits with cardboard and

metal caliper was taken. This was not stated, but probably could be safely assumed.

b. The population of the difference in head measurements between cardboard and metal calipers is normally distributed. To see if this is true, look at the histogram, number of outliers, and the normal probability plot. (If you wish, you can look at the normal probability plot first. If it doesn’t look linear, then you may want to look at the histogram and number of outliers at this point.)

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Figure #9.2.3: Histogram of Differences in Head Measurements

This histogram looks normal. There are no outliers in the difference data set.

Figure #9.2.4: Normal Probability Plot of Differences in Head Measurements

The probability plot on the differences looks somewhat linear.

So you can assume that the distribution of the difference in weights is normal. 4. Find the sample statistic, test statistic, and p-value

Using the TI-83/84, put x1 into L1 and x2 into L2. Then go onto the name L3, and type L1− L2 . The calculator will calculate the differences for you and put them in L3. Now go into STAT and move over to TESTS. Choose T- Test. The setup for the calculator is in figure #9.2.5. Figure #9.2.5: Setup for T-Test on Ti-83/84 Calculator

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Once you press ENTER on Calculate you will see the result shown in figure #9.2.6.

Figure #9.2.6: Results of T-Test on TI-83/84 Calculator

The t = 3.185 is the test statistic. The p = 0.0054147206 is the p-value. You also are given the sample mean of the differences, sample standard deviation of the differences, and sample size.

5. Conclusion Since the p-value < 0.05, reject Ho .

6. Interpretation There is enough evidence to show that the mean head measurements using the cardboard calipers are not the same as when using the metal calipers. So it looks like the New Zealand Air Force shouldn’t use the cardboard calipers.

Example #9.2.4: Confidence Interval for Paired Samples Using the Formula

A researcher wants to estimate the mean weight loss that people experience using a new program. She measures the weight of 6 randomly selected women before and after the weight loss program (see table #9.2.1). Find a 90% confidence interval for the mean the weight loss using the new program.

Solution: 1. State the random variables and the parameters in words.

These were stated in example #9.2.2, but are reproduced here for reference. x1 = weight of a woman after the weight loss program x2 = weight of a woman before the weight loss program µ1 = mean weight of a woman after the weight loss program µ2 = mean weight of a woman before the weight loss program

2. State and check the assumptions for the confidence interval

The assumptions were stated and checked in example #9.2.2.

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3. Find the sample statistic and confidence interval Sample Statistics:

From example #9.1.2 d = −21.5 sd = 3.15

The confidence level is 90%, so C = 90% There are six pairs, so the degrees of freedom are df = n −1= 6 −1= 5 Now look in table A.2. Go down the first column to 5, then over to the column headed with 90%. tc = 2.015 E = tc

sd n = 2.015 3.15

6 ≈ 2.6

d − E < µd < d + E −21.5 − 2.6 < µd < −21.5 + 2.6 −24.1 pounds < µd < −18.9 pounds

4. Statistical Interpretation: There is a 90% chance that

−24.1 pounds < µd < −18.9 pounds contains the true mean difference in weight loss.

5. Real World Interpretation: The mean weight loss is between 18.9 and 24.1

pounds. (Note, the negative signs tell you that the first mean is less than the second mean, and thus a weight loss in this case.)

Example #9.2.5: Confidence Interval for Paired Samples Using the Calculator

The New Zealand Air Force purchased a batch of flight helmets. They then found out that the helmets didn’t fit. In order to make sure that they order the correct size helmets, they measured the head size of recruits. To save money, they wanted to use cardboard calipers, but were not sure if they will be accurate enough. So they took 18 recruits and measured their heads with the cardboard calipers and also with metal calipers. The data in centimeters (cm) is in table #9.2.3 (“NZ helmet size,” 2013). Estimate the mean difference in measurements between the cardboard and metal calipers using a 95% confidence interval.

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Solution: 1. State the random variables and the parameters in words.

These were stated in example #9.2.3, but are reproduced here for reference. x1 = head measurement of recruit using cardboard caliper x2 = head measurement of recruit using metal caliper µ1 = mean head measurement of recruit using cardboard caliper µ2 = mean head measurement of recruit using metal caliper

2. State and check the assumptions for the hypothesis test

The assumptions were stated and checked in example #9.2.3.

3. Find the sample statistic and confidence interval Using the TI-83/84, put x1 into L1 and x2 into L2. Then go onto the name L3, and type L1− L2 . The calculator will now calculate the differences for you and put them in L3. Now go into STAT and move over to TESTS. Then chose TInterval. The setup for the calculator is in figure #9.2.11. Figure #9.2.11: Setup for TInterval on Ti-83/84 Calculator

Once you press ENTER on Calculate you will see the result shown in figure #9.2.12.

Figure #9.2.12: Results of TInterval on TI-83/84 Calculator

0.54 cm < µd < 2.68 cm

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4. Statistical Interpretation: There is a 95% chance that 0.54 cm < µd < 2.68 cm contains the true mean difference in head measurements between cardboard and metal calibers.

5. Real World Interpretation: The mean difference in head measurements

between the cardboard and metal calibers is between 0.54 and 2.68 cm. This means that the cardboard calibers measure on average the head of a recruit to be between 0.54 and 2.68 cm more in diameter than the metal calibers. That makes it seem that the cardboard calibers are not measuring the same as the metal calibers. (The positive values on the confidence interval imply that the first mean is higher than the second mean.)

Examples #9.2.2 and #9.2.4 use the same data set, but one is conducting a hypothesis test and the other is conducting a confidence interval. Notice that the hypothesis test’s conclusion was to reject Ho and say that there was a difference in the means, and the confidence interval does not contain the number 0. If the confidence interval did contain the number 0, then that would mean that the two means could be the same. Since the interval did not contain 0, then you could say that the means are different just as in the hypothesis test. This means that the hypothesis test and the confidence interval can produce the same interpretation. Do be careful though, you can run a hypothesis test with a particular significance level and a confidence interval with a confidence level that is not compatible with your significance level. This will mean that the conclusion from the confidence interval would not be the same as with a hypothesis test. So if you want to estimate the mean difference, then conduct a confidence interval. If you want to show that the means are different, then conduct a hypothesis test.

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Section 9.2: Homework In each problem show all steps of the hypothesis test or confidence interval. If some of the assumptions are not met, note that the results of the test or interval may not be correct and then continue the process of the hypothesis test or confidence interval. 1.) The cholesterol level of patients who had heart attacks was measured two days

after the heart attack and then again four days after the heart attack. The researchers want to see if the cholesterol level of patients who have heart attacks reduces as the time since their heart attack increases. The data is in table #9.2.4 (“Cholesterol levels after,” 2013). Do the data show that the mean cholesterol level of patients that have had a heart attack reduces as the time increases since their heart attack? Test at the 1% level. Table #9.2.4: Cholesterol Levels (in mg/dL) of Heart Attack Patients

Patient Cholesterol Level Day 2

Cholesterol Level Day 4

1 270 218 2 236 234 3 210 214 4 142 116 5 280 200 6 272 276 7 160 146 8 220 182 9 226 238 10 242 288 11 186 190 12 266 236 13 206 244 14 318 258 15 294 240 16 282 294 17 234 220 18 224 200 19 276 220 20 282 186 21 360 352 22 310 202 23 280 218 24 278 248 25 288 278 26 288 248 27 244 270 28 236 242

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2.) The cholesterol level of patients who had heart attacks was measured two days after the heart attack and then again four days after the heart attack. The researchers want to see if the cholesterol level of patients who have heart attacks reduces as the time since their heart attack increases. The data is in table #9.2.4 (“Cholesterol levels after,” 2013). Calculate a 98% confidence interval for the mean difference in cholesterol levels from day two to day four.

3.) All Fresh Seafood is a wholesale fish company based on the east coast of the U.S.

Catalina Offshore Products is a wholesale fish company based on the west coast of the U.S. Table #9.2.5 contains prices from both companies for specific fish types (“Seafood online,” 2013) (“Buy sushi grade,” 2013). Do the data provide enough evidence to show that a west coast fish wholesaler is more expensive than an east coast wholesaler? Test at the 5% level. Table #9.2.5: Wholesale Prices of Fish in Dollars

Fish

All Fresh Seafood Prices

Catalina Offshore Products Prices

Cod 19.99 17.99 Tilapi 6.00 13.99 Farmed Salmon 19.99 22.99 Organic Salmon 24.99 24.99 Grouper Fillet 29.99 19.99 Tuna 28.99 31.99 Swordfish 23.99 23.99 Sea Bass 32.99 23.99 Striped Bass 29.99 14.99

4.) All Fresh Seafood is a wholesale fish company based on the east coast of the U.S.

Catalina Offshore Products is a wholesale fish company based on the west coast of the U.S. Table #9.2.5 contains prices from both companies for specific fish types (“Seafood online,” 2013) (“Buy sushi grade,” 2013). Find a 95% confidence interval for the mean difference in wholesale price between the east coast and west coast suppliers.

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5.) The British Department of Transportation studied to see if people avoid driving on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations (“Friday the 13th,” 2013). The data for each location on the two different dates is in table #9.2.6. Do the data show that on average fewer people drive on Friday the 13th? Test at the 5% level. Table #9.2.6: Traffic Count

Dates 6th 13th 1990, July 139246 138548 1990, July 134012 132908 1991, September 137055 136018 1991, September 133732 131843 1991, December 123552 121641 1991, December 121139 118723 1992, March 128293 125532 1992, March 124631 120249 1992, November 124609 122770 1992, November 117584 117263

6.) The British Department of Transportation studied to see if people avoid driving

on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations (“Friday the 13th,” 2013). The data for each location on the two different dates is in table #9.2.6. Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level.

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7.) To determine if Reiki is an effective method for treating pain, a pilot study was carried out where a certified second-degree Reiki therapist provided treatment on volunteers. Pain was measured using a visual analogue scale (VAS) immediately before and after the Reiki treatment (Olson & Hanson, 1997). The data is in table #9.2.7. Do the data show that Reiki treatment reduces pain? Test at the 5% level. Table #9.2.7: Pain Measures Before and After Reiki Treatment

VAS before

VAS after

6 3 2 1 2 0 9 1 3 0 3 2 4 1 5 2 2 2 3 0 5 1 1 0 6 4 6 1 4 4 4 1 7 6 2 1 4 3 8 8

8.) To determine if Reiki is an effective method for treating pain, a pilot study was

carried out where a certified second-degree Reiki therapist provided treatment on volunteers. Pain was measured using a visual analogue scale (VAS) immediately before and after the Reiki treatment (Olson & Hanson, 1997). The data is in table #9.2.7. Compute a 90% confidence level for the mean difference in VAS score from before and after Reiki treatment.

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9.) The female labor force participation rates (FLFPR) of women in randomly selected countries in 1990 and latest years of the 1990s are in table #9.2.8 (Lim, 2002). Do the data show that the mean female labor force participation rate in 1990 is different from that in the latest years of the 1990s using a 5% level of significance? Table #9.2.8: Female Labor Force Participation Rates

Region and country

FLFPR 25-54 1990

FLFPR 25-54

Latest year of 1990s

Iran 22.6 12.5 Morocco 41.4 34.5 Qatar 42.3 46.5 Syrian Arab Republic 25.6 19.5 United Arab Emirates 36.4 39.7 Cape Verde 46.7 50.9 Ghana 89.8 90.0 Kenya 82.1 82.6 Lesotho 51.9 68.0 South Africa 54.7 61.7 Bangladesh 73.5 60.6 Malaysia 49.0 50.2 Mongolia 84.7 71.3 Myanmar 72.1 72.3 Argentina 36.8 54 Belize 28.8 42.5 Bolivia 27.3 69.8 Brazil 51.1 63.2 Colombia 57.4 72.7 Ecuador 33.5 64 Nicaragua 50.1 42.5 Uruguay 59.5 71.5 Albania 77.4 78.8 Uzbekistan 79.6 82.8

10.) The female labor force participation rates of women in randomly selected

countries in 1990 and latest years of the 1990s are in table #9.2.8 (Lim, 2002). Estimate the mean difference in the female labor force participation rate in 1990 to latest years of the 1990s using a 95% confidence level?

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11.) Table #9.2.9 contains pulse rates collected from males, who are non-smokers but do drink alcohol (“Pulse rates before,” 2013). The before pulse rate is before they exercised, and the after pulse rate was taken after the subject ran in place for one minute. Do the data indicate that the pulse rate before exercise is less than after exercise? Test at the 1% level. Table#9.2.9: Pulse Rate of Males Before and After Exercise

Pulse before

Pulse after

76 88 56 110 64 126 50 90 49 83 68 136 68 125 88 150 80 146 78 168 59 92 60 104 65 82 76 150

145 155 84 140 78 141 85 131 78 132

12.) Table #9.2.9 contains pulse rates collected from males, who are non-smokers but

do drink alcohol (“Pulse rates before,” 2013). The before pulse rate is before they exercised, and the after pulse rate was taken after the subject ran in place for one minute. Compute a 98% confidence interval for the mean difference in pulse rates from before and after exercise.

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Section 9.3 Independent Samples for Two Means This section will look at how to analyze when two samples are collected that are independent. As with all other hypothesis tests and confidence intervals, the process is the same though the formulas and assumptions are different. The only difference with the independent t-test, as opposed to the other tests that have been done, is that there are actually two different formulas to use depending on if a particular assumption is met or not. Hypothesis Test for Independent t-Test (2-Sample t-Test)

1. State the random variables and the parameters in words. x1 = random variable 1 x2 = random variable 2 µ1 = mean of random variable 1 µ2 = mean of random variable 2

2. State the null and alternative hypotheses and the level of significance The normal hypotheses would be Ho :µ1 = µ2 or Ho :µ1 − µ2 = 0 HA :µ1 < µ2 HA :µ1 > µ2 HA :µ1 ≠ µ2

HA :µ1 − µ2 < 0 HA :µ1 − µ2 > 0 HA :µ1 − µ2 ≠ 0

Also, state your α level here.

3. State and check the assumptions for the hypothesis test a. A random sample of size n1 is taken from population 1. A random sample of

size n2 is taken from population 2. Note: the samples do not need to be the same size, but the test is more robust if they are.

b. The two samples are independent. c. Population 1 is normally distributed. Population 2 is normally distributed.

Just as before, the t-test is fairly robust to the assumption if the sample size is large. This means that if this assumption isn’t met, but your sample sizes are quite large (over 30), then the results of the t-test are valid.

d. The population variances are equal, i.e. σ 1 2 =σ 2

2 . This assumption is no longer an assumption that most statisticians use. This is because it isn’t really realistic to assume that the variances are equal. So this assumption will not be checked.

4. Find the sample statistic, test statistic, and p-value

Sample Statistic: Calculate x1, x2, s1, s2, n1, n2

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Test Statistic: Since the assumption that σ 1

2 =σ 2 2 isn’t being satisfied, then

t = x1 − x2( )− µ1 − µ2( )

s1 2

n1 + s2

2

n2 Usually

µ1 − µ2 = 0 , since Ho :µ1 − µ2 = 0

Degrees of freedom: (the Welch–Satterthwaite equation)

df = A + B( ) 2

A2

n1 −1 + B

2

n2 −1

where A = s1 2

n1 and B = s2

2

n2

p-value: Use tcdf lower limit, upper limit, df( ) (Note: if HA :µ1 − µ2 < 0 , then lower limit is −1E99 and upper limit is your test statistic. If HA :µ1 − µ2 > 0 , then lower limit is your test statistic and the upper limit is 1E99 . If HA :µ1 − µ2 ≠ 0 , then find the p-value for HA :µ1 − µ2 < 0 , and multiply by 2.)

5. Conclusion

This is where you write reject Ho or fail to reject Ho . The rule is: if the p-value < α , then reject Ho . If the p-value ≥α , then fail to reject Ho

6. Interpretation

This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show HA is true, or you do not have enough evidence to show HA is true.

Confidence Interval for the Difference in Means from Two Independent Samples (2 Samp T-Int) The confidence interval for the difference in means has the same random variables and means and the same assumptions as the hypothesis test for independent samples. If you have already completed the hypothesis test, then you do not need to state them again. If you haven’t completed the hypothesis test, then state the random variables and means and state and check the assumptions before completing the confidence interval step.

1. Find the sample statistic and confidence interval Sample Statistic:

Calculate x1, x2, s1, s2, n1, n2 Confidence Interval: The confidence interval estimate of the difference µ1 − µ2 is

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Since the assumption that σ 1 2 =σ 2

2 isn’t being satisfied, then x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E

where E = tc s1

2

n1 + s2

2

n2

Degrees of freedom: (the Welch–Satterthwaite equation)

df = A + B( ) 2

A2

n1 −1 + B

2

n2 −1

where A = s1 2

n1 and B = s2

2

n2

2. Statistical Interpretation: In general this looks like, “there is a C% chance that

x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E contains the true mean difference.”

3. Real World Interpretation: This is where you state what interval contains the true difference in means, though often you state how much more (or less) the first mean is from the second mean.

The critical value is a value from the Student’s t-distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the difference in sample means, and the interval has a probability of containing the true difference in means, then you can think of this as the statement P x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E( ) = C . To find the critical value you use table A.2 in the Appendix. How to check the assumptions of two sample t-test and confidence interval: In order for the t-test or confidence interval to be valid, the assumptions of the test must be true. So whenever you run a t-test or confidence interval, you must make sure the assumptions are true. So you need to check them. Here is how you do this:

1. For the random sample assumption, describe how you took the two samples. Make sure your sampling technique is random for both samples.

2. For the independent assumption, describe how they are independent samples. 3. For the assumption about each population being normally distributed, remember

the process of assessing normality from chapter 6. Make sure you assess each sample separately.

4. You do not need to check the equal variance assumption since it is not being assumed.

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Example #9.3.1: Hypothesis Test for Two Means The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in table #9.3.1 (“Cholesterol levels after,” 2013). Do the data show that people who have had heart attacks have higher cholesterol levels over patients that have not had heart attacks? Test at the 1% level. Table #9.3.1: Cholesterol Levels in mg/dL

Cholesterol Level of Heart Attack Patients

Cholesterol Level of Healthy Individual

270 196 236 232 210 200 142 242 280 206 272 178 160 184 220 198 226 160 242 182 186 182 266 198 206 182 318 238 294 198 282 188 234 166 224 204 276 182 282 178 360 212 310 164 280 230 278 186 288 162 288 182 244 218 236 170

200 176

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Solution: 1. State the random variables and the parameters in words.

x1 = Cholesterol level of patients who had a heart attack x2 = Cholesterol level of healthy individuals µ1 = mean cholesterol level of patients who had a heart attack µ2 = mean cholesterol level of healthy individuals

2. State the null and alternative hypotheses and the level of significance The normal hypotheses would be Ho :µ1 = µ2 or Ho :µ1 − µ2 = 0 HA :µ1 > µ2 HA :µ1 − µ2 > 0 α = 0.01

3. State and check the assumptions for the hypothesis test a. A random sample of 28 cholesterol levels of patients who had a heart attack is

taken. A random sample of 30 cholesterol levels of healthy individuals is taken. The problem does not state if either sample was randomly selected. So this assumption may not be valid.

b. The two samples are independent since either they were dealing with patients who had heart attacks or healthy individuals.

c. Population of all cholesterol levels of patients who had a heart attack is normally distributed. Population of all cholesterol levels of healthy individuals is normally distributed. Patients who had heart attacks: Figure #9.3.1: Histogram of Cholesterol Levels of Patients who had Heart Attacks

This looks somewhat bell shaped. There are no outliers.

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Figure #9.3.2: Normal Probability Plot of Cholesterol Levels of Patients who had Heart Attacks

This looks somewhat linear. So, the population of all cholesterol levels of patients who had heart attacks is probably somewhat normally distributed. Healthy individuals: Figure #9.3.3: Histogram of Cholesterol Levels of Healthy Individuals

This does not look bell shaped. There are no outliers. Figure #9.3.4: Normal Probability Plot of Cholesterol Levels of Healthy Individuals

This doesn’t look linear. So, the population of all cholesterol levels of healthy individuals is probably not normally distributed.

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This assumption is not valid for the second sample. Since the sample is fairly large, and the t-test is robust, it may not be an issue. However, just realize that the conclusions of the test may not be valid.

4. Find the sample statistic, test statistic, and p-value

Sample Statistic: x1 ≈ 252.32, x2 ≈193.13, s1 ≈ 47.0642, s2 ≈ 22.3000, n1 = 28, n2 = 30 Test Statistic:

t = x1 − x2( )− µ1 − µ2( )

s1 2

n1 + s2

2

n2

= 252.32 −193.13( )− 0 47.06422

28 + 22.3000

2

30 ≈ 6.051

Degrees of freedom: (the Welch–Satterthwaite equation)

A = s1 2

n1 = 47.0642

2

28 ≈ 79.1085

B = s2 2

n2 = 22.3000

2

30 ≈16.5763

df = A + B( ) 2

A2

n1 −1 + B

2

n2 −1

= 79.1085 +16.5763( )2 79.10852

28 −1 + 16.5763

2

30 −1

≈ 37.9493

p-value: tcdf 6.051,1E99,37.9493( ) ≈ 2.44 ×10−7

Figure #9.3.5: Setup for 2-SampTTest on TI-83/84 Calculator

Note: the Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together.

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Figure #9.3.6: Results for 2-SampTTest on TI-83/84 Calculator

5. Conclusion Reject Ho since the p-value < α .

6. Interpretation

This is enough evidence to show that patients who have had heart attacks have higher cholesterol level on average from healthy individuals. (Though do realize that some of assumptions are not valid, so this interpretation may be invalid.)

Example #9.3.2: Confidence Interval for µ1 − µ2 The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in table #9.3.1 (“Cholesterol levels after,” 2013). Find a 99% confidence interval for the mean difference in cholesterol levels between heart attack patients and healthy individuals.

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Solution: 1. State the random variables and the parameters in words.

These were stated in example #9.3.1, but are reproduced here for reference. x1 = Cholesterol level of patients who had a heart attack x2 = Cholesterol level of healthy individuals µ1 = mean cholesterol level of patients who had a heart attack µ2 = mean cholesterol level of healthy individuals

2. State and check the assumptions for the hypothesis test The assumptions were stated and checked in example #9.3.1.

3. Find the sample statistic and confidence interval Sample Statistic: x1 ≈ 252.32, x2 ≈193.13, s1 ≈ 47.0642, s2 ≈ 22.3000, n1 = 28, n2 = 30 Test Statistic:

Degrees of freedom: (the Welch–Satterthwaite equation)

A = s1 2

n1 = 47.0642

2

28 ≈ 79.1085

B = s2 2

n2 = 22.3000

2

30 ≈16.5763

df = A + B( ) 2

A2

n1 −1 + B

2

n2 −1

= 79.1085 +16.5763( )2 79.10852

28 −1 + 16.5763

2

30 −1

≈ 37.9493

Since this df is not in the table, round to the nearest whole number. tc = 2.712

E = tc s1 2

n1 + s2

2

n2 = 2.712 47.0642

2

28 + 22.3000

2

30 ≈ 26.53

x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E 252.32 −193.13( )− 26.53< µ1 − µ2 < 252.32 −193.13( ) + 26.53

32.66 mg/dL < µ1 − µ2 < 85.72 mg/dL

Figure #9.3.7: Setup for 2-SampTInt on TI-83/84 Calculator

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Note: the Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together. Figure #9.3.8: Results for 2-SampTInt on TI-83/84 Calculator

4. Statistical Interpretation: There is a 99% chance that 32.66 mg/dL < µ1 − µ2 < 85.72 mg/dL contains the true difference in means.

5. Real World Interpretation: The mean cholesterol level for patients who had heart

attacks is anywhere from 32.66 mg/dL to 85,72 mg/dL more than the mean cholesterol level for healthy patients. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.)

If you do assume that the variances are equal, that is σ 1

2 =σ 2 2 , then the test statistic is:

t = x1 − x2( )− µ1 − µ2( ) sp

1 n1

+ 1 n2

where sp = n1 −1( )s12 + n2 −1( )s22 n1 −1( ) + n2 −1( )

sp = pooled standard deviation The Degrees of Freedom is: df = n1 + n2 − 2

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The confidence interval if you do assume that σ 1 2 =σ 2

2 has been met, is x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E

where E = tcsp 1 n1

+ 1 n2

and sp =

n1 −1( )s12 + n2 −1( )s22 n1 −1( ) + n2 −1( )

Degrees of Freedom: df = n1 + n2 − 2 tc is the critical value where C = 1−α

To show that the variances are equal, just show that the ratio of your sample variances is not unusual (probability is greater than 0.05). In other words, make sure the following is true.

P F > s1 2

s2 2

⎛ ⎝⎜

⎞ ⎠⎟ ≥ 0.05 (or P F > s2

2

s1 2

⎛ ⎝⎜

⎞ ⎠⎟ ≥ 0.05 so that the larger variance is in the

numerator). This probability is from an F-distribution. To find the probability on the TI-83/84 calculator use Fcdf s1

2 s2 2,1E99,n1 −1,n2 −1( ) .

Note: the F-distribution is very sensitive to the normal distribution. A better test for equal variances is Levene’s test, though it is more complicated. It is best to do Levene’s test when using statistical software (such as SPSS or Minitab) to perform the two-sample independent t-test.

Example #9.3.3: Hypothesis Test for Two Means

The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured (“SOCR 012708 id,” 2013). The data is in table #9.3.2. Is there enough evidence to show that beef has less sodium on average than poultry hotdogs? Use a 5% level of significance. Table #9.3.2: Hotdog Data

Sodium in Beef Hotdogs

Sodium in Poultry Hotdogs

495 430 477 375 425 396 322 383 482 387 587 542 370 359 322 357 479 528 375 513

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330 426 300 513 386 358 401 581 645 588 440 522 317 545 319 430 298 375 253 396

Solution: 1. State the random variables and the parameters in words.

x1 = sodium level in beef hotdogs x2 = sodium level in poultry hotdogs µ1 = mean sodium level in beef hotdogs µ2 = mean sodium level in poultry hotdogs

2. State the null and alternative hypotheses and the level of significance The normal hypotheses would be Ho :µ1 = µ2 or Ho :µ1 − µ2 = 0 HA :µ1 < µ2 HA :µ1 − µ2 < 0 α = 0.05

3. State and check the assumptions for the hypothesis test a. A random sample of 20 sodium levels in beef hotdogs is taken. A random

sample of 20 sodium levels in poultry hotdogs. The problem does not state if either sample was randomly selected. So this assumption may not be valid.

b. The two samples are independent since these are different types of hotdogs. c. Population of all sodium levels in beef hotdogs is normally distributed.

Population of all sodium levels in poultry hotdogs is normally distributed. Beef Hotdogs: Figure #9.3.9: Histogram of Sodium Levels in Beef Hotdogs

This does not look bell shaped. There are no outliers.

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Figure #9.3.10: Normal Probability Plot of Sodium Levels in Beef Hotdogs

This does not look linear. So, the population of all sodium levels in beef hotdogs is probably not normally distributed. Poultry Hotdogs: Figure #9.3.11: Histogram of Sodium Levels in Poultry Hotdogs

This does not look bell shaped. There are no outliers. Figure #9.3.12: Normal Probability Plot of Sodium Levels in Poultry Hotdogs

This does not look linear. So, the population of all sodium levels in poultry hotdogs is probably not normally distributed.

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This assumption is not valid. Since the samples are fairly large, and the t-test is robust, it may not be a large issue. However, just realize that the conclusions of the test may not be valid.

d. The population variances are equal, i.e. σ 1 2 =σ 2

2 . s1 ≈102.4347 s2 ≈ 81.1786 s1 2

s2 2 =

102.43472

81.17862 ≈1.592

Fcdf 1.592,1E99,19,19( ) ≈ 0.1597 ≥ 0.05 So you can say that these variances are equal.

4. Find the sample statistic, test statistic, and p-value Sample Statistic: x1 = 401.15, x2 = 450.2, s1 ≈102.4347, s2 ≈ 81.1786, n1 = 20, n2 = 20 Test Statistic: The assumption σ 1

2 =σ 2 2 has been met, so

sp = n1 −1( )s12 + n2 −1( )s22 n1 −1( ) + n2 −1( )

= 102.4347 2 *19 + 81.17862 *19

20 −1( ) + 20 −1( ) ≈ 92.4198

Though you should try to do the calculations in the problem so you don’t create round off error.

t = x1 − x2( )− µ1 − µ2( ) sp

1 n1

+ 1 n2

= 401.15 − 450.2( )− 0 92.4198 1

20 + 1 20

≈ −1.678 df = 20 + 20 − 2 = 38

p-value: Use tcdf −1E99,−1.678,38( ) ≈ 0.0508

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Figure #9.3.13: Setup for 2-SampTTest on TI-83/84 Calculator

Note: the Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes. Figure #9.3.14: Results for 2-SampTTest on TI-83/84 Calculator

5. Conclusion Fail to reject Ho since the p-value > α .

6. Interpretation

This is not enough evidence to show that beef hotdogs have less sodium than poultry hotdogs. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.)

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Example #9.3.4: Confidence Interval for µ1 − µ2 The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured (“SOCR 012708 id,” 2013). The data is in table #9.3.2. Find a 95% confidence interval for the mean difference in sodium levels between beef and poultry hotdogs.

Solution: 1. State the random variables and the parameters in words.

These were stated in example #9.3.1, but are reproduced here for reference. x1 = sodium level in beef hotdogs x2 = sodium level in poultry hotdogs µ1 = mean sodium level in beef hotdogs µ2 = mean sodium level in poultry hotdogs

2. State and check the assumptions for the hypothesis test The assumptions were stated and checked in example #9.3.3.

3. Find the sample statistic and confidence interval Sample Statistic: x1 = 401.15, x2 = 450.2, s1 ≈102.4347, s2 ≈ 81.1786, n1 = 20, n2 = 20 Confidence Interval: The confidence interval estimate of the difference µ1 − µ2 is

The assumption σ 1 2 =σ 2

2 has been met, so

sp = n1 −1( )s12 + n2 −1( )s22 n1 −1( ) + n2 −1( )

= 102.4347 2 *19 + 81.17862 *19

20 −1( ) + 20 −1( ) ≈ 92.4198

Though you should try to do the calculations in the formula for E so you don’t create round off error. df = n1 + n2 − 2 = 20 + 20 − 2 = 38 tc = 2.024

E = tcsp 1 n1

+ 1 n2

= 2.024 92.4198( ) 1 20

+ 1 20

≈ 59.15

x1 − x2( )− E < µ1 − µ2 < x1 − x2( ) + E 401.15 − 450.2( )− 59.15 < µ1 − µ2 < 401.15 − 450.2( ) + 59.15 −108.20 g < µ1 − µ2 <10.10 g

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Figure #9.3.15: Setup for 2-SampTInt on TI-83/84 Calculator

Note: the Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes. Figure #9.3.16: Results for 2-SampTInt on TI-83/84 Calculator

4. Statistical Interpretation: There is a 95% chance that −108.20 g < µ1 − µ2 <10.10 g contains the true difference in means.

5. Real World Interpretation: The mean sodium level of beef hotdogs is anywhere

from 108.20 g less than the mean sodium level of poultry hotdogs to 10.10 g more. (The negative sign on the lower limit implies that the first mean is less than

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the second mean. The positive sign on the upper limit implies that the first mean is greater than the second mean.)

Realize that many of assumptions are not valid in this example, so the interpretation may be invalid.

Section 9.3: Homework In each problem show all steps of the hypothesis test or confidence interval. If some of the assumptions are not met, note that the results of the test or interval may not be correct and then continue the process of the hypothesis test or confidence interval. Unless directed by your instructor, do not assume the variances are equal (except in problems 11 through 16). 1.) The income of males in each state of the United States, including the District of

Columbia and Puerto Rico, are given in table #9.3.3, and the income of females is given in table #9.3.4 (“Median income of,” 2013). Is there enough evidence to show that the mean income of males is more than of females? Test at the 1% level. Table #9.3.3: Data of Income for Males

\$42,951 \$52,379 \$42,544 \$37,488 \$49,281 \$50,987 \$60,705 \$50,411 \$66,760 \$40,951 \$43,902 \$45,494 \$41,528 \$50,746 \$45,183 \$43,624 \$43,993 \$41,612 \$46,313 \$43,944 \$56,708 \$60,264 \$50,053 \$50,580 \$40,202 \$43,146 \$41,635 \$42,182 \$41,803 \$53,033 \$60,568 \$41,037 \$50,388 \$41,950 \$44,660 \$46,176 \$41,420 \$45,976 \$47,956 \$22,529 \$48,842 \$41,464 \$40,285 \$41,309 \$43,160 \$47,573 \$44,057 \$52,805 \$53,046 \$42,125 \$46,214 \$51,630

Table #9.3.4: Data of Income for Females \$31,862 \$40,550 \$36,048 \$30,752 \$41,817 \$40,236 \$47,476 \$40,500 \$60,332 \$33,823 \$35,438 \$37,242 \$31,238 \$39,150 \$34,023 \$33,745 \$33,269 \$32,684 \$31,844 \$34,599 \$48,748 \$46,185 \$36,931 \$40,416 \$29,548 \$33,865 \$31,067 \$33,424 \$35,484 \$41,021 \$47,155 \$32,316 \$42,113 \$33,459 \$32,462 \$35,746 \$31,274 \$36,027 \$37,089 \$22,117 \$41,412 \$31,330 \$31,329 \$33,184 \$35,301 \$32,843 \$38,177 \$40,969 \$40,993 \$29,688 \$35,890 \$34,381

2.) The income of males in each state of the United States, including the District of

Columbia and Puerto Rico, are given in table #9.3.3, and the income of females is given in table #9.3.4 (“Median income of,” 2013). Compute a 99% confidence interval for the difference in incomes between males and females in the U.S.

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3.) A study was conducted that measured the total brain volume (TBV) (in mm3 ) of patients that had schizophrenia and patients that are considered normal. Table #9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV of schizophrenia patients (“SOCR data oct2009,” 2013). Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal? Test at the 10% level. Table #9.3.5: Total Brain Volume (in mm3 ) of Normal Patients

1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344 1319737 1688990 1292641 1512571 1635918

Table #9.3.6: Total Brain Volume (in mm3 ) of Schizophrenia Patients 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950 1563593 1317885 1420249 1363859 1238979 1286638 1325525 1588573 1476254 1648209 1354054 1354649 1636119

4.) A study was conducted that measured the total brain volume (TBV) (in mm3 ) of

patients that had schizophrenia and patients that are considered normal. Table #9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV of schizophrenia patients (“SOCR data oct2009,” 2013). Compute a 90% confidence interval for the difference in TBV of normal patients and patients with Schizophrenia.

5.) The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given

in table #9.3.7 and the lengths of NZ rivers that travel to the Tasman Sea are given in table #9.3.8 (“Length of NZ,” 2013). Do the data provide enough evidence to show on average that the rivers that travel to the Pacific Ocean are longer than the rivers that travel to the Tasman Sea? Use a 5% level of significance. Table #9.3.7: Lengths (in km) of NZ Rivers that Flow into the Pacific Ocean

209 48 169 138 64 97 161 95 145 90

121 80 56 64 209 64 72 288 322

Table #9.3.8: Lengths (in km) of NZ Rivers that Flow into the Tasman Sea 76 64 68 64 37 32 32 51 56 40 64 56 80 121 177 56 80 35 72 72 108 48

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6.) The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given in table #9.3.7 and the lengths of NZ rivers that travel to the Tasman Sea are given in table #9.3.8 (“Length of NZ,” 2013). Estimate the difference in mean lengths of rivers between rivers in NZ that travel to the Pacific Ocean and ones that travel to the Tasman Sea. Use a 95% confidence level.

7.) The number of cell phones per 100 residents in countries in Europe is given in

table #9.3.9 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 (“Population reference bureau,” 2013). Is there enough evidence to show that the mean number of cell phones in countries of Europe is more than in countries of the Americas? Test at the 1% level. Table #9.3.9: Number of Cell Phones per 100 Residents in Europe

100 76 100 130 75 84 112 84 138 133 118 134 126 188 129 93 64 128 124 122 109 121 127 152 96 63 99 95 151 147

123 95 67 67 118 125 110 115 140 115 141 77 98 102 102 112 118 118 54 23 121 126 47

Table #9.3.10: Number of Cell Phones per 100 Residents in the Americas 158 117 106 159 53 50 78 66 88 92 42 3

150 72 86 113 50 58 70 109 37 32 85 101 75 69 55 115 95 73 86 157 100 119 81 113 87 105 96

8.) The number of cell phones per 100 residents in countries in Europe is given in

table #9.3.9 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 (“Population reference bureau,” 2013). Find the 98% confidence interval for the different in mean number of cell phones per 100 residents in Europe and the Americas.

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9.) A vitamin K shot is given to infants soon after birth. Nurses at Northbay Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel (“SOCR data nips,” 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (table #9.3.11), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (table #9.3.12). Is there enough evidence to show that infants cried less on average when they are held by their mothers than if held using conventional methods? Test at the 5% level. Table #9.3.11: Crying Time of Infants Given Shots Using Conventional Methods

63 0 2 46 33 33 29 23 11 12 48 15 33 14 51 37 24 70 63 0 73 39 54 52 39 34 30 55 58 18

Table #9.3.12: Crying Time of Infants Given Shots Using New Methods 0 32 20 23 14 19

60 59 64 64 72 50 44 14 10 58 19 41 17 5 36 73 19 46 9 43 73 27 25 18

10.) A vitamin K shot is given to infants soon after birth. Nurses at Northbay

Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel (“SOCR data nips,” 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (table #9.3.11), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (table #9.3.12). Calculate a 95% confidence interval for the mean difference in mean crying time after being given a vitamin K shot between infants held using conventional methods and infants held by their mothers.

11.) Redo problem 1 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

12.) Redo problem 2 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

13.) Redo problem 7 testing for the assumption of equal variances and then use the

formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

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14.) Redo problem 8 testing for the assumption of equal variances and then use the

formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

15.) Redo problem 9 testing for the assumption of equal variances and then use the

formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

16.) Redo problem 10 testing for the assumption of equal variances and then use the

formula that utilizes the assumption of equal variances (follow the procedure in example 9.3.3).

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