Instructor-created question Question Help The manager of a paint supply store wa

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Instructor-created question Question Help The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturer's specifications that the standard deviation of the amount of paint is 0.02 gallon. You select a random sample of 55 cans, and the mean amount of paint per 1-gallon can is 0.996 gallon a. Is there evidence that th mean amount is different from 1.0 gallon? (Use = 0.01.) Let be the population mean. Determine the null hypothesis, Ho, and the alternative hypothesis, H1. Ho: = 1 H1: # 1 b. Compute the p-value and interpret its meaning. What is the p-value? 1380 (Round to four decimal places as needed.) Interpret the meaning of the p-value. Choose the correct answer below 0 A. Reject Ho . There is not sufficient evidence to warrant rejection of the claim that the mean amount is equal to 1 gallon. B. Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that the amount life is equal to 1 gallon C. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that the mean amount is equal to 1 gallon. D. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the mean amount is equal to 1 gallon. c. Construct a 95% confidence interval estimate of the population mean amount of paint per 1-gallon. 0.9890 1.0030 (Round to four decimal places as needed.) d. Compare the results of (a) and (c). What conclusions do you reach? 0 A. No meaningful conclusions can be reached. B. The results of (a) and (c) are not the same. yc. The results of (a) and (c) are the same

Explanation / Answer

Solution:- Given X = 0.996g, s = 0.02g n = 55

Z(0.025) = 1.96

95% confidence interval : X +/- Z*(s/sart(n))
: 0.996 +/- 1.96*(0.02/sqrt(55))
: 0.996 - 1.96*(0.02/sqrt(55)) , 0.996 + 1.96*(0.02/sqrt(55))
: (0.9907 , 1.0013)

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