In a survey conducted to determine the cost of vacations, 674 individuals were randomly sampled

Hypothesis Tests Suggested Problems

1. A machine that produces ball bearings is set so that the average diameter is .50 inch. In a sample of 100 ball bearings, it was found that x-bar = .51 inch. Assuming that the standard deviation is .05 inch, can we conclude at the 5% significance level that the mean diameter is not .50 inch? (Keller 5^th Ed. 10.33 p.333)

2. Past experience indicates that the monthly long-distance telephone bill is normally distributed with a mean of $17.85 and a standard deviation of $3.87. After an advertising campaign aimed at increasing long-distance telephone usage, a random sample of 25 household bills was taken. The results are listed in file (Data). (Keller 5^th Ed. 10.41 p.333)

a) Do the data allow us to infer at the 10% significance level that the campaign was successful?

b) What assumption must you make to answer part (a)?

3. A school-board administrator believes that the average number of days absent per year among students is less than 10 days. From past experience, he knows that the population standard deviation is 3 days. In testing to determine whether his belief is true, he could use either of the following plans. (Keller 5^th Ed. 10.57 p.342)

n = 100, a = .05
n = 100, a = .01
n = 250 , a = .05

Which plan has the lower probability of type II error, given that the true population average is 9 days. Why?

4. A diet doctor claims that the average North American is more than 20 pounds overweight. To test his claim, a random sample of 100 North Americans were weighed, and the difference between their actual weight and their ideal weight was calculated. The data are stored in file Data. Do these data allow us to infer at the 5% significance level that the doctor's claim is true? (Keller 5^th Ed. 11.32 p.363)

5. One important factory in inventory control is the variance of the daily demand for the product. An industrial engineer has developed the optimal order quantity and reorder point, assuming that the variance is equal to 250. Recently, the company has experienced some inventory problems, which induced the operations manager to doubt the assumption. To examine the problem, the manager took a sample of 25 daily demands recorded in file Data . Do these data provide sufficient evidence at the 5% significance level to infer that the Industrial Engineer's assumption about the variance is wrong? (Keller 5^th Ed. 11.46 p.373)

6. The January 3, 1993 edition of the Atlanta journal reported on a survey conducted in October of 1992. A random sample of 612 residents of Georgia were asked "Do you feel that things in Georgia are generally going in the right direction or do you feel things have gotten off the wrong path?" The possible responses are as follows:

Right track (1)

Wrong track (2)

No opinion (0)

The responses are stored in the frequency distribution below:

0 = 50

1 = 365

2 = 197

Estimate with 99% confidence the proportion of all Georgians who had an opinion and that opinion that things have gotten off on the wrong track. (Keller 11.30 p.409)

7. Barnes 9.11

8. Barnes 9.12

9. Barnes 9.19

10. Barnes 9.25

Hypothesis Tests Suggested Problems Solutions

1. (Keller 5^th Ed. 10.33 p.333)

.50

= = 2.00, p-value = 2P(z > 2.00) = .0456

Rejection region: z > or z <

Conclusion: Reject the null hypothesis. There is enough evidence to infer that the population mean diameter s not equal to .50.

2. (Keller 5^th Ed. 10.41 p.333)

a 17.85

17.85

= = 1.66, p-value = P(z > 1.66) = .0485

Rejection region: z >

Conclusion: Reject the null hypothesis. There is enough evidence to infer that the campaign was successful.

Excel Printout

Test of Hypothesis About MU (SIGMA Known)

Test of MU = 17.85 Vs MU greater than 17.85
SIGMA = 3.87
Sample mean = 19.1316
Test Statistic: z = 1.6558
P-Value = 0.0489

b The standard deviation after the campaign is assumed to be the same as before the campaign. That is,

= 3.85.

3. (Keller 5^th Ed. 10.57 p.342)

10.57 i Rejection region: <

< 9.30

= P( > 9.30 given = 9) = = P(z > 1) = .1587

Excel Printout

Probability of a Type II error
	Left-tail Test
H0: MU	10
SIGMA	3
Sample size	100
ALPHA	0.01
H1: MU	9

Critical value(s)	9.30
Prob(Type II error)	0.1570

i i Rejection region: <

< 9.43

= P( > 9.43 given = 9) = = P(z > 1.24) = .1075

Excel Printout

Probability of a Type II error
	Left-tail Test
H0: MU	10
SIGMA	3
Sample size	75
ALPHA	0.05
H1: MU	9

Critical value(s)	9.43
Prob(Type II error)	0.1071

iii Rejection region: <

< 9.46

= P( > 9.46 given = 9) = = P(z > 1.08) = .1401

Excel Printout

Probability of a Type II error
	Left-tail Test
H0: MU	10
SIGMA	3
Sample size	50
ALPHA	0.1
H1: MU	9

Critical value(s)	9.46
Prob(Type II error)	0.1411

4. (Keller 5^th Ed. 11.32 p.363)

= 20

> 20

= = 3.12

Rejection region: t > = 1.660

Conclusion: Reject the null hypothesis. There is enough evidence to infer that the doctor's claim is true.

Excel Printout

Test of Hypothesis About MU (SIGMA Unknown)

Test of MU = 20 Vs MU greater than 20
Sample standard deviation = 13.4222
Sample mean = 24.19
Test Statistic: t = 3.1217
P-Value = 0.0012

5. (Keller 5^th Ed. 11.46 p.373)

= 250

250

= = 25.98, p-value (Excel) = .7088

Rejection region: = 39.3641 or == 12.4011

Conclusion: Do not reject the null hypothesis. There is not sufficient evidence to infer that the variance is not equal to 250.

Excel Printout

Chi-squared Test of a Variance

Sample variance	270.58
Sample size	25

Hypothesized variance	250

Chi-squared Stat	25.98
Two-tail p-value	0.7088