# UNIT 3 SUCCESS GUIDE

SUCCESS GUIDE

1 | GB 513 Unit 3 Success Guide v.6.13.17

UNIT 3 SUCCESS GUIDE

This unit is the other “most difficult” one. Hypothesis testing has two parts: setting-up the hypotheses and calculating the critical values to determine results. They both pose difficulty for a lot of students. The seminar will be on the first and the recorded lecture will be on the second. You need to make sure you understand both, otherwise you will not be able to get to the right conclusions.

1. As always, start by reading the chapters and studying the solved examples.

2. Watch the lecture video in document sharing. It focuses on why we do

hypothesis testing, how to do it with Excel and solves two sample problems.

3. Watch this from Khan Academy:

values

This one talks more about how to write the null and alternative hypotheses

(which a lot of students get wrong) and also solves the problem using

formulas.

4. Watch the sample problem solutions in Course Resources.

5. If you still want more videos, search YouTube for “hypothesis testing.” Several

introductory level videos are available, such as

Email your instructor if you find any of these links to be broken.

Avoid these mistakes!

GENERAL NOTES

RESOURCES

COMMON MISTAKES IN THE ASSIGNMENT

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 Students commonly get the null and alternative hypotheses reversed, or

get them completely wrong.

 Students also commonly do not state the hypothesis fully. This is correct:

“null hypothesis: there is no difference between the average salary for

group 1 and the average salary of group 2.” This is not sufficient: “ho:

x1=x2”

 Students sometimes compare the averages of the two groups and base

their determination on which one is greater, rather than properly doing a

hypothesis test.

 Students sometimes do the calculations correctly, but do not write out

what the conclusion is. This is correct: “We therefore reject the null

hypothesis, which means we conclude that there is a difference between

the average salaries of the two groups.” This is not sufficient: “reject null.”

 In the last problem, students run the test but then have no idea which

metric to use in order to make a conclusion.

The questions below are very similar to what you need to solve in the assignment.

Some, but not all, of these solutions were demonstrated on video and recorded

for the live binder by the math tutors.

SAMPLE PROBLEM 1 FOR ASSIGNMENT PROBLEM 4

A hole-punch machine is set to punch a hole 1.84 centimeters in diameter in a

strip of sheet metal in a manufacturing process. The strip of metal is then creased

and sent on to the next phase of production, where a metal rod is slipped

through the hole. It is important that the hole be punched to the specified

diameter of 1.84 cm. To test punching accuracy, technicians have randomly

sampled 12 punched holes and measured the diameters. The data (in

centimeters) follow. Use an alpha of .10 to determine whether the holes are being

punched an average of 1.84 centimeters.

Assume the punched holes are normally distributed in the population.

1.81 1.89 1.86 1.83 1.85 1.82 1.87 1.85

1.84 1.86 1.88 1.85

SAMPLE PROBLEMS AND SOLUTIONS

SOLUTION

n = 12 x = 1.85083 s = .02353 df = 12 – 1 = 11  = .10

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x  

n

1.85083  1.84

.02353

12

x  

n

8.37  8.3

.1895

20

SAMPLE PROBLEM 2 FOR ASSIGNMENT PROBLEM 4

The following data (in pounds), which were selected randomly from a normally

distributed population of values, represent measurements of a machine part that

is supposed to weigh, on average, 8.3 pounds.

8.1 8.4 8.3 8.2 8.5 8.6 8.4 8.3 8.4 8.2

8.8 8.2 8.2 8.3 8.1 8.3 8.4 8.5 8.5 8.7

Use these data and alpha =0.01 to test the hypothesis that the parts average 8.3

pounds.

SAMPLE PROBLEM 1 FOR ASSIGNMENT PROBLEM 5

H0: µ = 1.84

Ha: µ  1.84

For a two-tailed test, /2 = .05 critical t.05,11 = 1.796

t =  = 1.59

Since t = 1.59 < t11,.05 = 1.796,

The decision is to fail to reject the null hypothesis

SOLUTION

n = 20 x = 8.37

Ho: µ = 8.3

s = .1895 df = 20-1 = 19  = .01

Ha: µ  8.3

For two-tail test, /2 = .005 critical t.005,19 = ±2.861

t =  = 1.65

Observed t = 1.65 < t.005,19 = 2.861

The decision is to Fail to reject the null hypothesis

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p̂  p 

.4316  .48

p  q

n

(.48)(.52)

380

p̂  p 

.3333  .31

p  q

n

(.31)(.69)

600

The Independent Insurance Agents of America conducted a survey of insurance

consumers and discovered that 48% of them always reread their insurance

policies, 29% sometimes do, 16% rarely do, and 7% never do. Suppose a large

insurance company invests considerable time and money in rewriting policies so

that they will be more attractive and easy to read and understand. After using

the new policies for a year, company managers want to determine whether

rewriting the policies significantly changed the proportion of policyholders who

always reread their insurance policy. They contact 380 of the company’s

insurance consumers who purchased a policy in the past year and ask them

whether they always reread their insurance policies. One hundred and sixty-four

respond that they do. Use a 1% level of significance to test the hypothesis.

SAMPLE PROBLEM 2 FOR ASSIGNMENT PROBLEM 5

A survey was undertaken by Bruskin/Goldring Research for Quicken to determine

how people plan to meet their financial goals in the next year. Respondents were

allowed to select more than one way to meet their goals. Thirty-one percent said

that they were using a financial planner to help them meet their goals. Twenty-

four percent were using family/friends to help them meet their financial goals

followed by broker/accountant (19%), computer software (17%), and books

(14%). Suppose another researcher takes a similar survey of 600 people to test

these results. If 200 people respond that they are going to use a financial planner

to help them meet their goals, is this proportion enough evidence to reject the

31% figure generated in the Bruskin/Goldring survey using

If 158 respond that they are going to use family/friends to help them meet their

financial goals, is this result enough evidence to declare that the proportion is

significantly higher than Bruskin/Goldring’s figure of .24 if alpha = 0.05?

SOLUTION

Ho: p = .48

Ha: p  .48

n = 380 x = 164  = .01 /2 = .005 z.005 = +2.575

= .4316

n 380

z = = -1.89

Since the observed z = -1.89 is greater than z.005= -2.575, the decision is to

fail to reject the null hypothesis. There is not enough evidence to declare that

the proportion is any different than .48.

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p  q

(.24)(.76)

600

SAMPLE PROBLEM FOR ASSIGNMENT PROBLEM 6

Since the Assignment requires you to use the data analysis tool pack for this

problem, the best way to prepare is to watch the lecture video, which gives two

examples, both solved using Excel.

SOLUTION

Ho: p = .31

Ha: p  .31

n = 600 x = 200  = .10 /2 = .05 z.005 = +1.645

= .3333

z = = 1.23

Since the observed z = 1.23 is less than z.005= 1.645, the decision is to fail to reject

the null hypothesis. There is not enough evidence to declare that the proportion

is any different than .31.

Ho: p = .24

Ha: p > .24

n = 600 x = 158  = .05 z.05 = 1.645

n 600

Since the observed z = 1.34 is less than z.05= 1.645, the decision is to fail to reject

the null hypothesis. There is not enough evidence to declare that the

proportion is less than .24.

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