The dataset StudentSurvey has information from males and females on the number o

Question

The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given Descriptive Statistics: TV Minimum 03 Maximum Mean 5.237 7.620 6.427 0.000 3.0005.000 10.000 40.000 Variable Gender StDev Median 169 4.100 0.000 2.5004 6.000 20.000 TV 192 Click here for the dataset associated with this question. (a) In the sample, which group watches more TV, on average? By how much? Enter the exact answer watch more TV by hours per week. exact number, no tolerance LINK TO TEXT b) Use the summary statistics to compute a 99% confidence interval or the difference in mean number o hours spent exercising 1 where by all female students at this university and represents the number of hours spent watching TV per week by all male students at this university Round your answers to two decimal places. represents the number of hours spent watching TV per week The 99% confidence interval is to

Explanation / Answer

Correct Answer: option (B) Same up to round-off error

Since First we calculate F - Test, is there any significance difference between the variances in two population

From the given data

Test Statistic, F: 0.4066

Lower Critical F: 0.7745469
Upper Critical F: 1.288057

Here F -value is not in critical value,

So, we conclude that there is significance difference between two population variances

Now, we should conduct t-test for independence of mean with unequal variance

The given results are same to t-test for indindependence of mean with unequal variance results

99% Confidence interval:
-3.833678 < µ1-µ2 < -0.9263219

P-Value: 0.0000

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