9.25 A manufacturer of chocolate candies uses machines to package candies as the
ID: 3200129 • Letter: 9
Question
9.25
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 pack-ages 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 in a particular sample of 50 packages, the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce.
a. Is there evidence that the population mean amount is different from 8.17 ounces? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning.
Explanation / Answer
Null hypothesis: = 8.17
Alternative hypothesis: 8.17
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
= 0.051 / sqrt(50)
= 0.007
DF = n - 1 = 50 - 1 = 49
t = (x - ) / SE
= (8.159 - 8.17)/ 0.007
= -1.57
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.57
Thus, the P-value = 0.122
a) There is not sufficient evidence that the population mean amount is different from 8.17 ounces
b) Interpret results. Since the P-value (0.122) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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