Your manager 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 gallons. You select a random

sample of 50 cans, and the mean amount of paint per 1-gallon can is 0.995. pwzmean and pwciesigmaknown.xls

1. Is there evidence that the mean amount is different from 1.0 gallon (use ? = 0.01)?

a. State the null and alternative hypothesis

b. State the decision rule

c. Compute the Z-test statistic

d. State the decision

2. Compute the p-value and interpret the meaning.

3. Construct a 99% confidence interval estimate of the population mean amount of paint.

4. Based on your findings, what conclusions do you reach? Present your findings and interpret the meaning. How is this data and information useful for effective managerial decision-making?nd the mean amount of paint per 1-gallon can is 0.995.

SIGMA KNOWN:

Estimate for the Mean Paper Length Population Standard Deviation 0.02 Sample Mean 10.998 Sample Size 100 Confidence Level 9543 Intermediate Calculations Standard Error of the Mean 0.002 z value -1.9800 Interval Half Width 0.0039 Confidence Interval Interval Lower Limit 10.9841 Interval Upper Limit 11.0019Estimate for the Mean Paper Length Data Population Star 0.02 Sample Mean 10.998 Sample Size 100 Confidence Lev Intermediate Calculations Standard Error 0.002 anJBA Z -1.9600 0.0039 Confidence Interval JEMOT BABIUI 10.9941 Interval Upper 11.0019t Test for the Hypothesis of the Mean Data Null Hypothesis 120 Level of Significance 0.05 Sample Size 12 Sample Mean 112.85 Sample Standard Deviation 20 8 Intermediate Calculations Standard Error of the Mean 6.0044 Degrees of Freedom 11 c Test Stationc -1.1908 Two-Tail Test Lower Critical Value -2.2010 Upper Critical Value 2.2010 p-Value 0.2586 Do not reject the null hypothesisCOMPUTE_LOWER t Test for the Hypothesis of the Mean Data Null Hypothesis 158.77 Level of Significance 0.05 Sample Size 25 Sample Mean 147 9 Sample Standard Deviation 20 Intermediate Calculations Standard Error of the Mean 4.0000 Degrees of Freedom 24 Test Statistic 2.6175 Lower-Tail Test Lower Critical Value -1.7109 p-Value 0.0046 Reject the null hypothesis Calculations Area For one-tailed tests TDIST value 0.004767 1-TDIST value 0.995239COMPUTE_UPPER Test for the Hypothesis of the Mean Data Null Hypothesis 158.77 Level of significance 0.05 Sample Size 25 sample Mean 147 5 Sample Standard Deviation 20 Intermediate Calculations Standard Error of the Mean 4.0000 Degrees of Freedom 24 Test Statistic 2.6175 Upper-Tail Test Upper Critical Value 1.7109 p-Value 0.9852 Do not reject the null hypothesis Calculations Area For one-tailed tests TDIST value 0.004767 1-TDIST value 0.995239COMPUTE_ALL t Test for the Hypothesis of the Mean Data Null Hypothesis 120 Level of significance 0.05 Sample Size 12 Sample Mean 112.85 Sample Standard Deviation 20.8 Intermediate Calculations Standard Error of the Mean 6.0044 Degrees of Freedom 11 t Test Statistic -1.1906 Two-Tail Test Lower Critical Value -2.2010 Upper Critical Value 2.2010 p-Value 0.2568 Do not reject the null hypothesis Calculations Area Lower-Tail Test For one-tailed tests Lower Critical Value -1.7959 TDIST value 0.1294 p-Value 0.1294 1-TDIST value 0.8706 Do not reject the null hypothesis Upper-Tail Test Upper Critical Value 1.7859 p-Value 0.6706 Do not reject the null hypothesisCOMPUTE_ALL FORM It Test for the Hypothesis of the Mean Data Null Hypothesis 120 Level of Significance 0.05 Sample Size 12 Sample Mean 112 85 Sample Standard Deviation 20.8 Intermediate Calculations Standard Error of the Mean 6.0044 Degrees of Freedom 11 f Test Statistic -1.190B Two-Tail Test Lower Critical Value -2.2010 Upper Critical Value 2.2010 p-Value 0.2568 Do not reject the null hypothesis Calculations Area Lower-Tail Test For one-tailed tests: Lower Critical Value -1.7959 TDISTV| 0 12940017 p-Value 0.1294 1-TDIST 0.87059989 Do not reject the null hypothesis Upper-Tail Test Upper Critical Value 1.7858 p-Value 0.6706 Do not reject the null hypothesis