The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the population mean life of a large shipment of CFLs is equal to 7,497 hours. The population standard deviation is 1,080 hours. A random sample of 81 light bulbs indicates a sample mean life of 7,197 hours. a. Let μ be the population mean. Determine the null hypothesis, H0, and the alternative hypothesis, H1. A. H0 : μ = 7,197 and H1 : μ ≠ 7,197, because the sample mean is always used in hypothesis testing B. H0 : μ = 7,497 and H1 : X = 7,197, because the population mean is used in H0 and the sample mean in H1 C. H0 : μ ≠ 7,497 and H1 : μ = 7,497, because H0 never uses the equal symbol D. H0 : μ = 7,497 and H1 : μ ≠ 7,497 because the "goal" or "historical data" is always used when stating the hypotheses What is the value of the test statistic? (Round to two decimal places to the right of the decimal point as needed.) See fig 1. What is/are the critical value(s)? (Round to two decimal places to the right of the decimal point as needed. Use a comma to separate answers as needed.) A. -1.96 and +1.96 found using±NORM.S.INV(0.05/2) B.-1.96 found using (NORM.S.INV(0.05/2) C.+1.96 found using -(NORM.S.INV(0.05/2)) D.-1.64 and +1.64 found using±NORM.S.INV(0.05) What is the final conclusion? A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. B. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. C. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. D. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. b. What is the p-value? (Round to three decimal places to the right of the decimal point as needed.) Interpret the meaning of the p-value. Choose the correct answer below. A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. C. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. D. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. (Round to one decimal place as needed.) A. 6,961.8 ≤ μ ≤ 7,432.2, found using 7,197± CONFIDENCE.NORM(0.05,1080,81) B. 7,261.8 ≤ μ ≤ 7,732.2, found using 7,497±CONFIDENCE.NORM(0.05,1080,81) C. 6,928.0 ≤ μ ≤ 7,466.0, found using 7,197±CONFIDENCE.NORM(0.05/2,1080,81) D. 7,170.9 ≤ μ ≤ 7,223.1, found using 7,197±CONFIDENCE.NORM(0.05,120,81) d. Compare the results of (a) and (c). What conclusions do you reach? A.The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours. B.The results of (a) and (c) are not the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours. C.The results of (a) and (c) are the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours. D.The results of (a) and (c) are the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours

The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the population mean life of a large shipment of CFLs is equal to 7,497 hours. The population standard deviation is 1,080 hours. A random sample of 81 light bulbs indicates a sample mean life of 7,197 hours. a. Let μ be the population mean. Determine the null hypothesis, H0, and the alternative hypothesis, H1. A. H0 : μ = 7,197 and H1 : μ ≠ 7,197, because the sample mean is always used in hypothesis testing B. H0 : μ = 7,497 and H1 : X = 7,197, because the population mean is used in H0 and the sample mean in H1 C. H0 : μ ≠ 7,497 and H1 : μ = 7,497, because H0 never uses the equal symbol D. H0 : μ = 7,497 and H1 : μ ≠ 7,497 because the "goal" or "historical data" is always used when stating the hypotheses What is the value of the test statistic? (Round to two decimal places to the right of the decimal point as needed.) See fig 1. What is/are the critical value(s)? (Round to two decimal places to the right of the decimal point as needed. Use a comma to separate answers as needed.) A. -1.96 and +1.96 found using±NORM.S.INV(0.05/2) B.-1.96 found using (NORM.S.INV(0.05/2) C.+1.96 found using -(NORM.S.INV(0.05/2)) D.-1.64 and +1.64 found using±NORM.S.INV(0.05) What is the final conclusion? A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. B. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. C. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. D. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. b. What is the p-value? (Round to three decimal places to the right of the decimal point as needed.) Interpret the meaning of the p-value. Choose the correct answer below. A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours. C. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. D. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours. c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. (Round to one decimal place as needed.) A. 6,961.8 ≤ μ ≤ 7,432.2, found using 7,197± CONFIDENCE.NORM(0.05,1080,81) B. 7,261.8 ≤ μ ≤ 7,732.2, found using 7,497±CONFIDENCE.NORM(0.05,1080,81) C. 6,928.0 ≤ μ ≤ 7,466.0, found using 7,197±CONFIDENCE.NORM(0.05/2,1080,81) D. 7,170.9 ≤ μ ≤ 7,223.1, found using 7,197±CONFIDENCE.NORM(0.05,120,81) d. Compare the results of (a) and (c). What conclusions do you reach? A.The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours. B.The results of (a) and (c) are not the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours. C.The results of (a) and (c) are the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours. D.The results of (a) and (c) are the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Related questions

Question

a. Let μ be the population mean. Determine the null hypothesis, H0, and the alternative hypothesis, H1.

A. H0 : μ = 7,197 and H1 : μ ≠ 7,197, because the sample mean is always used in hypothesis testing

B. H0 : μ = 7,497 and H1 : X = 7,197, because the population mean is used in H0 and the sample mean in H1

C. H0 : μ ≠ 7,497 and H1 : μ = 7,497, because H0 never uses the equal symbol

D. H0 : μ = 7,497 and H1 : μ ≠ 7,497 because the "goal" or "historical data" is always used when stating the hypotheses

What is the value of the test statistic? (Round to two decimal places to the right of the decimal point as needed.)

See fig 1.

What is/are the critical value(s)? (Round to two decimal places to the right of the decimal point as needed. Use a comma to separate answers as needed.)

A. -1.96 and +1.96 found using±NORM.S.INV(0.05/2)

B.-1.96 found using (NORM.S.INV(0.05/2)

C.+1.96 found using -(NORM.S.INV(0.05/2))

D.-1.64 and +1.64 found using±NORM.S.INV(0.05)

What is the final conclusion?

A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours.

B. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours.

C. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours.

D. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours.

b. What is the p-value? (Round to three decimal places to the right of the decimal point as needed.)

Interpret the meaning of the p-value. Choose the correct answer below.

A. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours.

B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,497 hours.

C. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours.

D. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,497 hours.

c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. (Round to one decimal place as needed.)

A. 6,961.8 ≤ μ ≤ 7,432.2, found using 7,197± CONFIDENCE.NORM(0.05,1080,81)

B. 7,261.8 ≤ μ ≤ 7,732.2, found using 7,497±CONFIDENCE.NORM(0.05,1080,81)

C. 6,928.0 ≤ μ ≤ 7,466.0, found using 7,197±CONFIDENCE.NORM(0.05/2,1080,81)

D. 7,170.9 ≤ μ ≤ 7,223.1, found using 7,197±CONFIDENCE.NORM(0.05,120,81)

d. Compare the results of (a) and (c). What conclusions do you reach?

A.The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours.

B.The results of (a) and (c) are not the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours.

C.The results of (a) and (c) are the same: there is not sufficient evidence to prove that the mean life is different from 7,497 hours.

D.The results of (a) and (c) are the same: there is sufficient evidence to prove that the mean life is different from 7,497 hours.

A.
B.
O C.
O D.
ZSTAT
ZSTAT
ZSTAT
ZSTAT
= +2.50, found by
=
=
=
+0.28, found by
- 2.50, found by
-0.28, found by
(7497 - 7197)
1080
√81
(7497 - 7197)
(1080)
(7197 - 7497)
1080
√81
(7197 - 7497)
(1080)
= + 2.50
= +0.28
=
- 2.50
= -0.28

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