Question

17. A tire manufacturer claims that the average life of a certain grade of tire is 50,000 miles or more when used under normal driving conditions. A random sample of 20 tires is selected and tested and the sample mean is 49,650 miles. Assume the population of the lives is to be normal with standard deviation of 5,000 miles. At the 5% significance level, the manufacturer’s claim is tested. What is the decision rule in this test?

A. Reject H₀ if Z_STAT > +1.96; otherwise, do not reject H₀.

B. Reject H₀ if Z_STAT < –1.645; otherwise, do not reject H₀.

C. Reject H₀ if t_STAT > –1.7291; otherwise, do not reject H₀.

D. Reject H₀ if t_STAT < –2.0930 or t_STAT > +2.0930; otherwise, do not reject H₀.

18. The manager of a gasoline supply corporation wants to estimate the actual amount of gasoline contained in standard 42-gallon barrels purchased from a nationally known manufacturer. The manufacturer’s specifications state that the amount of gasoline is normally distributed with a standard deviation of 0.65 gallon. A random sample of 25 barrels is selected, and the sample mean amount of gasoline per 42-gallon barrel is 42.15 gallons. Construct a 90% confidence interval estimate for the population mean amount of gasoline contained in a standard 42-gallon barrel.

A. 42.15 ± 1.7109 (0.65) / 25

B. 42 ± 1.645 (0.65) / 5

C. 42 ± 1.7109 (0.65) / 25

D. 42.15 ± 1.645 (0.65) / 5

19. “Is live lecture over Zoom an effective way to conduct online lectures?” A pilot sample of 50 students of UM reveals that 18 of them gave a positive response. If a researcher wants to estimate the proportion of positive responses from all the students of UM, what should be the minimum sample size used if she wants the width of a 95% confidence interval estimate to be 0.10?

A. 355

B. 97

C. 385

D. 89

20. A dean of a business school is interested in determining whether the mean grade point average (GPA) of students is less than3.04. The population standard deviation is 0.41 A random sample of 200 students indicates a sample mean GPA of 2.94. A test is conducted at the 0.05 level of significance to determine whether the mean grade point average (GPA) of students is less than 3.04. What is the test statistic value in this test?

A. +0.244

B. –3.449

C. –0.244

D. +3.449

21.A random sample of 40 bags of flour is weighed. It is found that the sample mean is 453.08 grams. All weights of the bags of flour have a population standard deviation of 5.42 grams. What is the lower limit of an 80% confidence interval for the population mean weight?

A. 452.405 grams

B. 452.360 grams

C. 451.983 grams

D. No solution

Answer 1

Q17: x̅ = 49650, σ = 5000, n = 20  

Null and Alternative hypothesis:  

Ho : µ ≥ 50000 ; H1 : µ < 50000  

Critical value, z crit = NORM.S.INV(0.05) =    -1.645

Reject Ho if z < -1.645  

Test statistic:  

z = (x̅- µ)/(σ/√n) = (49650 - 50000)/(5000/√20) =    -0.3130

Decision: Do not reject the null hypothesis  

Answer: B. Reject H₀ if Z_STAT < –1.645; otherwise, do not reject H₀.

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Q18: x̅ = 42.15, σ = 0.65, n = 25

90% Confidence interval :  

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) =    1.645

Lower Bound = x̅ - z_c*σ/√n = 42.15 - 1.645 * 0.65/√25 = 41.9362

Upper Bound = x̅ + z_c*σ/√n = 42.15 + 1.645 * 0.65/√25 = 42.3638

Answer: D. 42.15 ± 1.645 (0.65) / 5

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Q19: Proportion, p = 18/50 = 0.36      

Margin of error, E = 0.10/2 = 0.05      

Confidence Level, CL = 0.95      

Significance level, α = 1 - CL = 0.05      

Critical value, z = NORM.S.INV(0.05/2) = 1.96   

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.36 * 0.64)/ 0.05²          

= 354.0288 = 355

Answer: A. 355

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Q20: Test statistic:  

z = (x̅- µ)/(σ/√n) = (2.94 - 3.04)/(0.41/√200) = -3.449

Answer: B. -3.449

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Q21: x̅ = 453.08, σ = 5.42, n = 40  

80% Confidence interval :  

At α = 0.2 two tailed critical value, z_c = ABS(NORM.S.INV(0.2/2)) =    1.282

Lower Bound = x̅ - z_c*σ/√n = 453.08 - 1.282 * 5.42/√40 = 451.983

Answer: C. 451.983 grams