# Testing Procedures and Five-Step Hypothesis Testing

Need Help with the following exercises:

I have attached the five-step hypothesis testing procedure

14. The null and alternate hypotheses are:

H0: 1 2

H1: 1 2

A random sample of 15 observations from the first population revealed a sample mean of

350 and a sample standard deviation of 12. A random sample of 17 observations from the

second population revealed a sample mean of 342 and a sample standard deviation of 15.

At the .10 significance level, is there a difference in the population means?

Note: Use the five-step hypothesis testing procedure for the following exercises.

17. Ms. Lisa Monnin is the budget director for Nexus Media, Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information.

Sales ($) 131 135 146 165 136 142

Audit ($) 130 102 129 143 149 120 139

At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value?

Note: Use the five-step hypothesis testing procedure to solve the following exercises.

21. The management of Discount Furniture, a chain of discount furniture stores in the Northeast,

designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople

were selected at random, and their weekly incomes before and after the plan were

recorded.

Salesperson Before After

Sid Mahone $320 $340

Carol Quick 290 285

Tom Jackson 421 475

Andy Jones 510 510

Jean Sloan 210 210

Jack Walker 402 500

Peg Mancuso 625 631

Anita Loma 560 560

John Cuso 360 365

Carl Utz 431 431

A. S. Kushner 506 525

Fern Lawton 505 619

Was there a significant increase in the typical salesperson's weekly income due to the innovative

incentive plan? Use the .05 significance level. Estimate the p-value, and interpret it.

43. Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different

advertising approaches, they use different media to reach potential buyers. The mean

annual family income for 75 people making inquiries at the first development is $150,000,

with a standard deviation of $40,000. A corresponding sample of 120 people at the second

development had a mean of $180,000, with a standard deviation of $30,000. At the .05 significance

level, can Fairfield conclude that the population means are different?

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#### Solution Preview

Please see the attached file.

14. The null and alternate hypotheses are:

H0: 1 2

H1: 1 2

A random sample of 15 observations from the first population revealed a sample mean of

350 and a sample standard deviation of 12. A random sample of 17 observations from the

second population revealed a sample mean of 342 and a sample standard deviation of 15.

At the .10 significance level, is there a difference in the population means?

Note: Use the five-step hypothesis testing procedure for the following exercises.

1) Hypothesis:

2) Level of Significance:

3) Test Statistics:

Where s =

=

= 13.6821

= 1.65056

4) Critical Region:

Â± 1.697

5) Conclusion:

Since the computed test statistics is less than the critical region, we fail to reject the null hypothesis. At the 10% level of significance, the data does not provide sufficient evidence that the populations' means are different.

17. Ms. Lisa Monnin is the budget director for Nexus Media, Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information.

Sales ($) 131 135 146 165 136 142

Audit ($) 130 102 129 143 149 120 139

At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value?

Sales

($) Audit

($)

131 130

135 102

146 129

165 143

136 149

142 120

139

...

#### Solution Summary

This solution contains step-by-step calculations that conduct hypothesis testing on each scenario. A null and alternative hypothesis is provided and the test statistic is calculated and compared to the p-value to make a decision in either accepting or rejecting the null hypothesis.