Question

The pass completion percentages of 10 college football quarterbacks for their freshman and sophomore seasons are shown in the table below. At a =( completion percentages have changed? Complete parts (a) through (e) below. Player 1 1 2 3 4 5 6 7 8 9 10 - age 67.5 62.2 56.9 60.0 62.5 52.0 56.0 62.9 54.9 57.2 (freshman) Completion percentage 67.1 56.2 63.660.6 60.8 60.5 61.9 61.956.4 59.1 (sophomore) Oc.tk (c) Calculated and so Calculated d = -1.600 (Round to three decimal places as needed.) Calculate d. SA = 4.379 (Round to three decimal places as needed.) (d) Find the standardized test statistic t. t = - 1.16 (Round to two decimal places as needed.) (e) Decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim. the null hypothesis. There enough evidence to the claim that the pass completion percentages have

Answer 1

Sample 1 Sample 2 Difference = Sample 1 - Sample 2 67.5 67.1 0.40000000000 001 62.2 56.2 56.9 63.6 -6.7 60 60.6 -0.6 62.5 60.8 1.7 52 60.5 -8.5 56 61.9 -5.9 62.9 61.9 54.9 56.4 -1.5 57.2 59.1 -1.9 Average 59.21 60.81 -1.6 St. Dev. 4.63 3.223 4.379 10 In 10 From the sample data, it is found that the corresponding sample means are: X1 = 59.21 X2 = 60.81 Also, the provided sample standard deviations are: $1 = 4.63 $2 = 3.223 and the sample size is n = 10. For the score differences we have ✓ = -1.6 SD = 4.379

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Md = 0 Ha: up 70 This corresponds to a two-tailed test, for which a t-test for two paired samples be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 9. Hence, it is found that the critical value for this two-tailed test is te = 2.262, for a = 0.05 and df = 9. The rejection region for this two-tailed test is R = {t: 1t| > 2.262}. (3) Test Statistics The t-statistic is computed as shown in the following formula: D t=- -1.6 4.379/V10 = -1.155 sp/n (4) Decision about the null hypothesis Since it is observed that |t| = 1.155 < te = 2.262, it is then concluded that the null hypothesis is not rejected. (5)conclusion Based on above data we failed to reject null hypothesis so we can say that we don't have enough evidence to support our claim that completion percentage have changed

t Table cum. probl one-tail two-tails t50 0.50 1.00 €75 0.25 0.50 t.30 0.20 0.40 1.85 0.15 0.30 t go 0.10 0.20 t.95 0.05 0.10 € 975 0.025 0.05 t go 0.01 0.02 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 9 | 0.000 0.000 0.000 0.000 0.000 0.000 151 0.000 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 .093 1.088 1.083 1.079 1.076 1.074 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 .345 1.341 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 1 1