The first pic is the question. Since some tutors have been solving the problems wrong recently (no judgment), I’ve added a sample question to solve problem. Recreational Time A researcher wishes to see if there is a difference between the mean number of hours per week thata family with no children participates in recreational activities and a family with children participates in recreationalactivities. She selects two random samples and the data are shown. Use u, for the mean number of families with nochildren. At a = 0.05, is there a difference between the means? Use the critical value method and tables.No children8.82.535Children10.12.935Send data to ExcelPart: 0 / 5Part 1 of 5(a) State the hypotheses and identify the claim.Ho:(Choose one) ▼H:|(Choose one) ▼This hypothesis test is a (Choose one) v test. Sample QuestionRecreational Time A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates inrecreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. Use u, for the meannumber of families with no children. At a =0.01, is there a difference between the means? Use the critical value method and tables.No children8.92.634Children10.82.834Send data to Excel(a) State the hypotheses and identify the claim.(b) Find the critical values.(c) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim.The null hypothesis is the statement that there is no difference between the means. This is equivalent to H: H, =H.The alternative hypothesis is the statement that there is a difference between the means. This is equivalent to H,: H, #µy.The problem asks, "Are the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates inrecreational activities different?" Hence, the claim is the alternative hypothesis.(b) Find the critical values.From the figure, for a two-tailed test with a= 0.01, the critical values are z=12.58.a= 0.10, C.V.=±1.65¥ = ri : °Ha= 0.05, C.V.-±1.96H :H#ka- 0.01, C.V.-£2.58(c) Compute the test value.Using the formula for the z test for comparing two means from independent populations, compute the test value:(X,- X2)- (H, – H2)(8.9 – 10.8) – 02.62 2.823434=-2.90Hence, the test value rounded to 2 decimal places is z= - 2.90.(d) Make the decision.Since the test value does fall in the critical region, reject the null hypothesis.-2.90 -2.582.58(e) Summarize the results.

Question

The first pic is the question. Since some tutors have been solving the problems wrong recently (no judgment), I’ve added a sample question to solve problem. Recreational Time A researcher wishes to see if there is a difference between the mean number of hours per week thata family with no children participates in recreational activities and a family with children participates in recreationalactivities. She selects two random samples and the data are shown. Use u, for the mean number of families with nochildren. At a = 0.05, is there a difference between the means? Use the critical value method and tables.No children8.82.535Children10.12.935Send data to ExcelPart: 0 / 5Part 1 of 5(a) State the hypotheses and identify the claim.Ho:(Choose one) ▼H:|(Choose one) ▼This hypothesis test is a (Choose one) v test. Sample QuestionRecreational Time A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates inrecreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. Use u, for the meannumber of families with no children. At a =0.01, is there a difference between the means? Use the critical value method and tables.No children8.92.634Children10.82.834Send data to Excel(a) State the hypotheses and identify the claim.(b) Find the critical values.(c) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim.The null hypothesis is the statement that there is no difference between the means. This is equivalent to H: H, =H.The alternative hypothesis is the statement that there is a difference between the means. This is equivalent to H,: H, #µy.The problem asks, &#34;Are the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates inrecreational activities different?&#34; Hence, the claim is the alternative hypothesis.(b) Find the critical values.From the figure, for a two-tailed test with a= 0.01, the critical values are z=12.58.a= 0.10, C.V.=±1.65¥ = ri : °Ha= 0.05, C.V.-±1.96H :H#ka- 0.01, C.V.-£2.58(c) Compute the test value.Using the formula for the z test for comparing two means from independent populations, compute the test value:(X,- X2)- (H, – H2)(8.9 – 10.8) – 02.62 2.823434=-2.90Hence, the test value rounded to 2 decimal places is z= - 2.90.(d) Make the decision.Since the test value does fall in the critical region, reject the null hypothesis.-2.90 -2.582.58(e) Summarize the results.

Accepted Answer

Given that 
A researcher wishes to see if there is a difference between the mean number of hours per…