For the procedure, the options are on the second screenshot. The 2002 general Social Survey asked a large number of people how much time they spent watching TVeach day. The mean number of hours was 4.16 with a standard deviation of 2.23. Assume that in a sampleof 52 teenagers, the sample standard deviation of daily TV time is 1.98 hours, and that the population ofTV watching times is normally distributed. Under 10% significance level can you conclude that thepopulation standard deviation of TV watching times for teenagers is different from 2.23?Procedure: Select an answerAssumption Select an answerO PopuO The One proportion Z Hypothesis Testbonses are both greater than 10O PopuOne variance x Hypothesis TestO SimpO Norm One mean T Hypothesis TestO SamdOne mean Z Hypothesis TestStep 1. Hypue opwhere ? v is theHp: Select an answer v =Select an answervand the units are? vHa: Select an answer v?vand the test is Select an answer vStep 2. The significance level a =Step 3. Compute the value of the test statistic: Select an answer v(Round the answer to 3decimal places)Step 4. Testing Procedure: (Round the answers to 3 decimal places)CVAPVACompute the P-value of the teststatistic:Provide the critical value(s) for the Rejection Region:left CV isand right CV isP-value isStep 5. Decision:CVAPVAIs the test statistic in the rejection region?Is the P-value less than the significance level?Conclusion: Select an answerStep 6. Interpretation: The 2002 general Social Survey asked a large number of people how much time they spent watching TVeach day. The mean number of hours was 4.16 with a standard deviation of 2.23. Assume that in a sampleof 52 teenagers, the sample standard deviation of daily TV time is 1.98 hours, and that the population ofTV watching times is normally distributed. Under 10% significance level can you conclude that thepopulation standard deviation of TV watching times for teenagers is different from 2.23?Procedure: Select an answerAssumptions: (select everything that applies)O Population standard deviation is knownO The number of positive and negative responses are both greater than 10O Population standard deviation is unknownO Simple random sampleO Normal populationO Sample size is greater than 30Step 1. Hypotheses Set-Up:where ? v is theSelect an answerHọ: Select an answer v=vand the units areH: [Select an answer vand the test is Select an answer vStep 2. The significance level a =56Step 3. Compute the value of the test statistic: Select an answer v=(Round the answer to 3decimal places)Step 4. Testing Procedure: (Round the answers to 3 decimal places)CVAPVACompute the P.value of the testProvide the critical value(s) for the Rejection Region:statistic::left CV isand right CV isP-value isStep 5. Decision:CVAPVAIs the test statistic in the rejection region?Is the P-value less than the significance level?Conclusion: [Select an answerStep 6. Interpretation:

Name: For the procedure, the options are on the second screenshot. The 2002
Uploaded: 2024-03-20T06:38:44.000Z
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Description: For the procedure, the options are on the second screenshot. The 2002 general Social Survey asked a large number of people how much time they spent watching TVeach day. The mean number of hours was 4.16 with a standard deviation of 2.23. Assume that

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Accepted Answer

here we use 
χ2  one variance test 
given mean 4.16
standard deviation 2.23