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 packages 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 8.168 ounces,with a sample standard deviation of 0.051 ounce.Complete parts(a)and(b). Click here to view page 1 of the critical values for the t Distribution. The critical value(s) is(are) (Round to four decimal places as needed.
Question
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 packages 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 8.168 ounces,with a sample standard deviation of 0.051 ounce.Complete parts(a)and(b). Click here to view page 1 of the critical values for the t Distribution. The critical value(s) is(are) (Round to four decimal places as needed.
Solution
(a) Hypothesis Test:
Null Hypothesis (H0): μ = 8.17 ounces (The mean amount packaged is 8.17 ounces) Alternative Hypothesis (H1): μ ≠ 8.17 ounces (The mean amount packaged is not 8.17 ounces)
(b) Test Statistic Calculation:
We can use the formula for the t statistic in a one-sample t test:
t = (x̄ - μ) / (s/√n)
where: x̄ = sample mean = 8.168 ounces μ = population mean = 8.17 ounces s = sample standard deviation = 0.051 ounces n = sample size = 50
Substituting these values into the formula, we get:
t = (8.168 - 8.17) / (0.051/√50) = -0.28
This is the test statistic.
(c) Critical Value:
The critical value for a two-tailed test with 49 degrees of freedom (n-1) and a significance level of 0.05 can be found in the t-distribution table or calculated using a statistical software or calculator. The critical value is approximately ±2.0096.
(d) Conclusion:
Since the calculated t statistic (-0.28) is within the range of the critical values (-2.0096 to 2.0096), we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean amount packaged is different from 8.17 ounces.
Similar Questions
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 packages 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 deviation of 0.059 ounce.Complete parts (a)and(b) State the null and alternative hypotheses. H0μ= 8.17 H1:μ≠8.17 Determine the test statistic. The test statistic is
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 8.171 ounces,with a sample standard deviation of 0.053 ounce.Complete parts (a)and(b) Click here to view page 1 of the critical values for the t Distribution. Click here to view page 2 of the critical values for the t Distribution. 回 a.Is there evidence that the population mean amount is different from 8.17 ounces?(Use a 0.10 level of significance. State the null and alternative hypotheses.
An automatic coffee machine dispenses cups of coffee whose volume per cup varies normally with the mean μ = 10 oz. A quality-control researcher randomly selects 8 cups of coffee from the machine and finds that in this sample the mean volume is 9.92 oz. and the standard deviation is 0.23 oz.We want to determine if these data provide enough evidence to conclude that the mean volume per cup is below the target level.
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