A manufacturer of chocolate candies uses machines to package candies as they move along a filling line.

Although the packages are labeled as 8 ounces, the company wants the package to contain a mean of 8.17

ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is

selected periodically, and the packaging process is stopped if there is evidence that the mean amount

packaged is different from 8.17 ounces. Suppose that a particular sample of 50 packages, the mean

amount dispersed is 8.159 ounces with a sample standard deviation of 0.051 ounces. Use the pwtmean file to help you.

1. Is there evidence that the population mean amount is different from 8.17 ounces? Use an ? = 0.05

level of significance.

a. State the null and alternative hypotheses

b. State the decision rule

c. Compute the test statistic

d. State the decision

2. Determine the p-value and interpret its meaning

3. Based on your findings, what conclusions do you reach? How is this data useful for effective

managerial decision-making?

t Test for the Hypothesis of the Mean Data Null Hypothesis 120 Level of Significance 0.05 Sample Size Sample Mean 112.85 Sample Standard Deviation 20.8 Intermediate Calculations Standard Error of the Mean 8.0044 Degrees of Freedom 11 -1. 1808 Two-Tall Test Lower Critical Value -2.2010 Upper Critical Value 2.2010 p-Value 0.2568 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 5 Sample Standard Deviation 20 Intermediate Calculations Standard Error of the Mean 4.0000 Degrees of Freedom 24 Test Statistic -2.8175 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 t 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 Intermediate Calculations Standard Error of the Mean 4.0000 Degrees of Freedom 24 t 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 0.895239COMPUTE_ALL t Test for the Hypothesis of the Mean Data Null Hypothesis 120 Level of Significance 0.09 Sample Size 12 Sample Mean 112.85 Sample Standard Deviation 20.6 Intermediate Calculations Standard Error of the Mean 6.0044 Degrees of Freedom 11 t Test Statistic -1.1908 Two-Tail Test Lower Critical Value 2.2010 Upper Critical Value 2.2010 p-Value 0.2586 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 1.7858 p-Value 0.6706 Do not reject the null hypothesisCOMPUTE_ALL_FORMULAS 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 20.6 Intermediate Calculations Standard Error of the Mean 6.0044 Degrees of Freedom 11 Test Statistic -1.1908 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 D-Value 0.1294 1-TDIST 0.87059983 Do not reject the null hypothesis Upper-Tail Test Upper Critical Value 1.7858 p-Value 0.6706 Do not reject the null hypothesis