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.

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

The waiting time to check out of a supermarket has had a population mean of 11.02 minutes.Recently,in an effort to reduce the waiting time,the supermarket has experimented with a system in which there is a single waiting line with multiple checkout servers.A sample of 50 customers was selected,and their mean waiting time to check out was 9.79 minutes with a sample standard deviation of 5.6 minutes.Complete parts(a)through(d). 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.At the 0.01 level of significance,using the critical value approach to hypothesis testing,is there evidence that the population mean waiting time to check out is less than 11.02 minutes? What are the null and alternative hypotheses for this test?

The restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits: 48 45 44 43 46 47 42 46 47 45 47 49 Note: The data appears to be approximately normally distributed. Test the bartender's ability to pour 45 mL at the 5% level of significance. T-Distribution Table a. Calculate the sample mean and standard deviation. x̄ = 45.750 Round to three decimal places if necessary s= 0.000 Round to three decimal places if necessary b. Calculate the test statistic. t= 0.000 Round to three decimal places if necessary

The waiting time to check out of a supermarket has had a population mean of 8.49 minutes.Recently,in an effort to reduce the waiting time,the supermarket has experimented with a system in deviation of 4.9 minutes.Complete parts(a)through(d). 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.At the 0.01 level of significance,using the critical value approach to hypothesis testing,is there evidence that the population mean waiting time to check out is less than 8.49 minutes?H0:μ≥8.49 H1:μ<8.49 What is the test statistic for this​ test?