Answer to Question #314329 in Statistics and Probability for Ann

Question

Conduct Hypothesis Testing Using Small-Sample Tests.

An experimental study was conducted by a researcher to determine if new time slot has an effect on the performance of students in mathematics. Ten randomly selected students participated in the study. Toward the end of the investigation, a standardized assessment was conducted. The sample mean was X=75 and the standard deviation s=20. In the standardization of the test, the mean was 65 and the standard deviation was 10. Based on the evidence at hand, is the new time slot effective? Use α=0.05.

Answer 1

$H_0:\mu _1=\mu _2\\H_1:\mu _1>\mu _2\\n_1=10\\n_2=10\\\bar{x}_1=75\\\bar{x}_2=65\\s_1=20\\s_2=10\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{20^2}{10}+\frac{10^2}{10} \right) ^2}{\frac{1}{9}\left( \frac{20^2}{10} \right) ^2+\frac{1}{9}\left( \frac{10^2}{10} \right) ^2}=13.2353\approx 13\\T=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}}=\frac{75-65}{\sqrt{\frac{20^2}{10}+\frac{10^2}{10}}}=1.41421\\P-value:\\P\left( T>1.41421 \right) =F_{t,13}\left( 1.41421 \right) =0.0904$

Since the P-value is greater than the significance level, the new time slot is not effective.