Suppose we are interested in comparing female versus male class attendance. The following data is provided: ROUND TO FOUR DECIMAL PLACES Gender Female Male Sample Size 10 11 Average Attendance Percentage 88.5 86.75 Sample Standard Deviation 12.53 13.07 a. What is the 99% confidence interval for the difference between the average attendance between female and male? b. What is the test statistic for the difference for female and male? c. What is the p-value?

Sample sizes for female and male are , n1 = 10. n2 = 11. Sample means for female and male are , x1 = 88.5 x2 = 86.75 Sample standard deviations for female and male are , s1 = 12.53 s2 = 13.07. It is asked to find The 99% confidence interval for the difference between the average attendance between female and male. The test statistic for the difference for female and male. The p-value. To find confidence interval and test statistic, t we have two formula's ; 1) Un pooled : In this variances are not equal. 2) Pooled : In this variances are equal. So to find confidence interval and test statistic , t first we have to check either variances are equal or not. For that we need to set hypotheses. Ho : σ1² = σ2² Ha : σ1² ≠ σ2² Since we've not equal sign in alternative hypotheses it is two tailed test. Find F-value : The formula to calculate F value is as follows; F = (larger standard deviation )²(smaller standard deviation)²=(13.07)²(12.53)²=1.0880. Find Critical value : To find the critical value we need to know the following things: 1) Degrees of freedom for numerator , df1 = n1-1 = 10-1 = 9. 2) Degrees of freedom for denominator , df2= n2-1 = 11-1 = 10. 3) It is a two-tailed test. 4) Given level of confidence , c = 99% so this implies Significance level , α = 1 - c = 1 - 0.99 = 0.01 We can easily find this critical value by using MS-Excel. The Excel function to find critical value is =FINV ( α , df1 , df2) Plug values and calculate critical value. That is , =FINV(0.01 , 9 ,10) Thus , critical value = 4.9424. Decision and conclusion : The decision rule says that , If F-value > Critical value, reject the null hypothesis. If F-value < Critical value, fail to reject the null hypothesis. Here our calculated F-value = 1.0880 and Critical value = 4.9424 that is F-value < critical value so we fail to reject the null hypotheses. Decision: Fail to reject the null hypothesis. Conclusion : Here variance are equal so we will use "POOLED" formula. That is , use pooled formula to find confidence interval and test statistic , t.It is asked to construct, 99% confidence interval for the difference between the average attendance between female and male. That is given level of confidence , c = 99% or 0.99 (in decimal form). We can find this confidence interval by using TI-84 calculator. To find this please refer the following steps; Press the STAT button and highlight TESTS. Scroll down to 0 : 2-Sample T Interval. Highlight Stats and enter the asked values it will look like , Highlight Calculate and hit ENTER. It returns the confidence interval in interval notation: (min, max). Thus , the 99% confidence interval for the difference between the average attendance between female and male is (-14.27 , 17.772).It is asked to find the the test statistic for the difference for female and male and P-value. For this firstly we need to set the hypotheses. Ho : µ1 - µ2 = 0 Ha : µ1 - µ2 ≠0. Since we have not equal sign in alternative hypotheses so it is two tailed test. We can find this test statistic and p-value by using TI-84 calculator. To find this please refer the following steps; Press the STAT button and highlight TESTS. Scroll down to 4 : 2-Sample T Test. Highlight Stats and enter the asked values it will look like , Highlight Calculate and hit ENTER. It returns the test statistic, t and p-value. Thus from this output , Test statistic , t = 0.3124904668 till 4 decimal places it would be 0.3125. P-value = 0.7580715576 till 4 decimal places it would be 0.7581.Part a) The 99% confidence interval for the difference between the average attendance between female and male is (-14.27 , 17.772). Part b) The test statistic for the difference for female and male is 0.3125. Part c) The p-value is 0.7581.