The manager of a paint supply store wants to determine whether the mean amount o

Question

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 specifications that the standard deviation of the amount of paint is 0.02 gallon. You select a random sample of 50 cans, and the mean amount of paint per 1-gallon can is 0.996 gallon. Complete parts (a) through (c) below. Let ? be the population mean. Determine the null hypothesis, Ho, and the alternative hypothesis, H1 What is the test statistic? ZSTATRound to two decimal places as needed.) What is/are the critical value(s)? (Use ?= 0.01.) (Round to two decimal places as needed. Use a comma to separate answers as needed.) What is the final conclusion? O A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the mean amount is different from 1.0 gallon O B. Fail to reject Ho. There is not sufficient evidence that the mean amount is different from 1.0 gallon ° C. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that the mean amount is different from 1.0 gallon ( D. Reject Ho-There is not sufficient evidence to warrant rejection of the claim that the mean amount is different from 1.0 gallon. b. Construct a 99% confidence interval estimate of the population mean amount of paint per 1-gallon. SHS (Round to four decimal places as needed.) c. Compare the results of (a) and (b). What conclusions do you reach? O A. The results of (a) and (b) are the same O B. The results of (a) and (b) are not the same ° C. No meaningful I conclusions can be reached.

Explanation / Answer

The statistical software output for this problem is:

One sample Z summary hypothesis test:
? : Mean of population
H0 : ? = 1
HA : ? ? 1
Standard deviation = 0.02

Hypothesis test results:

Hence,

Hypotheses:

H0 : ? = 1
HA : ? ? 1

Test statistic = -1.41

Critical values = -2.58, 2.58

Final conclusion: Option B is correct.

b) 99% confidence interval: 0.9887 < ? < 1.0033

c) Option A is correct.

Mean n Sample Mean Std. Err. Z-Stat P-value ? 50 0.996 0.0028284271 -1.4142136 0.1573

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