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 8.171 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. 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 H0:μ=8.17 H1:μ≠8.17 (Type integers or decimals. Identify the critical value(s). The critical value(s)is(are) (Round to four decimal places as needed.Use a comma to separate answers 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 8.171 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. 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 H0:μ=8.17 H1:μ≠8.17 (Type integers or decimals. Identify the critical value(s). The critical value(s)is(are) (Round to four decimal places as needed.Use a comma to separate answers as needed.)
Solution
To answer this question, we first need to state the null and alternative hypotheses.
The null hypothesis (H0) is that the population mean (μ) is equal to 8.17 ounces. The alternative hypothesis (H1) is that the population mean (μ) is not equal to 8.17 ounces.
Next, we need to identify the critical value(s).
To do this, we use the t-distribution table. Since we are using a 0.10 level of significance and the sample size is 50, we have 49 degrees of freedom (n-1).
Looking at the t-distribution table, the critical value for a two-tailed test with 49 degrees of freedom at a 0.10 level of significance is approximately ±1.676.
So, the critical value(s) are -1.676 and 1.676.
Now, we need to calculate the test statistic.
The test statistic is calculated as follows:
t = (sample mean - population mean) / (sample standard deviation / sqrt(n))
t = (8.171 - 8.17) / (0.053 / sqrt(50))
t = 0.001 / 0.0075
t = 0.133
Since the calculated test statistic (0.133) is within the critical region (-1.676, 1.676), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to suggest that the population mean amount is different from 8.17 ounces.
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Context: 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 0.059 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. 。○ State the null and alternative hypotheses H0μ=8.17 H1μ≠8.17 (Type integers or decimals.) Identify the critical value(s). The critical value(s)is(are)-2.6800,2.6800 (Round to four decimal places as needed.Use a comma to separate answers as needed.) Determine the test statistic. Determine the p-value and interpret its meaning. The p-value is
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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