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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

Question

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

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Solution

(a) The null hypothesis (H0) is that the mean amount of candies in the packages is 8.17 ounces (μ=8.17). The alternative hypothesis (H1) is that the mean amount of candies in the packages is not 8.17 ounces (μ≠8.17).

(b) The critical values are -2.6800 and 2.6800. These are the values beyond which we would reject the null hypothesis.

(c) To determine the test statistic, we would use the formula for the t-statistic, which is (sample mean - population mean) / (sample standard deviation / sqrt(sample size)). In this case, the sample mean is 8.168 ounces, the population mean is 8.17 ounces, the sample standard deviation is 0.059 ounces, and the sample size is 50.

(d) The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. To interpret its meaning, if the p-value is less than the significance level (usually 0.05), we would reject the null hypothesis and conclude that there is evidence that the mean amount of candies in the packages is different from 8.17 ounces. If the p-value is greater than the significance level, we would not reject the null hypothesis and conclude that there is not enough evidence to say that the mean amount of candies in the packages is different from 8.17 ounces.

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Similar Questions

ough 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. 0· a.Is there evidence that the population mean amount is different from 8.17 ounces?(Use a 0.01 level of significance.) State the null and alternative hypotheses. H0:μ=8.17 H1μ≠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.) ChatGPT To test whether there is evidence that the population mean amount is different from 8.17 ounces at a 0.01 level of significance, you can set up the null and alternative hypotheses as follows: Null Hypothesis (H0): The population mean amount of candies is equal to 8.17 ounces. H0: μ = 8.17 Alternative Hypothesis (H1): The population mean amount of candies is different from 8.17 ounces. H1: μ ≠ 8.17 To identify the critical value(s) for a two-tailed test at a 0.01 level of significance, you can use a t-distribution table or a calculator. Since the sample size is 50, you have 49 degrees of freedom (df = n - 1). Using a t-distribution table with 49 degrees of freedom and a significance level of 0.01 (two-tailed), you can find the critical values. Typically, for a two-tailed test at 0.01 significance level, the critical values are approximately ±2.6851 (rounded to four decimal places). So, the critical value(s) for this test are approximately ±2.6851. If the calculated t-statistic falls beyond these critical values in either the lower or upper tail, you would reject the null hypothesis.

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.

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.Below is the output:Which of the following represents the correct conclusion we can make based on the output (and at the usual significance level of 0.05)? The data provide enough evidence to reject H0 and to conclude that the mean volume per cup is lower than the target level of 10 oz. The data provide enough evidence to accept H0 and to conclude that the mean volume per cup is at the target level of 10 oz. The data do not provide enough evidence to reject H0, so we accept it, and conclude that the mean volume per cup is at the target level of 10 oz. The data do not provide enough evidence to reject H0, nor to conclude that the mean volume per cup is lower than the target level of 10 oz.Question 4

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