in an effort to improve scores on college - entrance exams, the principal at a large high school chooses 35…

in an effort to improve scores on college - entrance exams, the principal at a large high school chooses 35 students at random who have already taken the exam. the students are enrolled in a six - week prep course to improve their scores. after the prep course, the students take the college - entrance exam again. the mean difference (first - second) in exam scores is - 1.75 points with a standard deviation of 7.12 points. assuming the conditions for inference are met, what is the p - value for testing the hypotheses $h_0:mu_{diff}=0;h_a:mu_{diff}>0$? find the t - table here. 0.0775 0.1551 0.8449 0.9225

in an effort to improve scores on college - entrance exams, the principal at a large high school chooses 35 students at random who have already taken the exam. the students are enrolled in a six - week prep course to improve their scores. after the prep course, the students take the college - entrance exam again. the mean difference (first - second) in exam scores is - 1.75 points with a standard deviation of 7.12 points. assuming the conditions for inference are met, what is the p - value for testing the hypotheses $h_0:mu_{diff}=0;h_a:mu_{diff}>0$? find the t - table here. 0.0775 0.1551 0.8449 0.9225

Answer

Explanation:

Step1: Calculate the t - statistic

The formula for the t - statistic in a paired - samples t - test is $t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$, where $\bar{d}$ is the sample mean difference, $\mu_d$ is the hypothesized population mean difference under the null hypothesis, $s_d$ is the sample standard deviation of the differences, and $n$ is the sample size. Here, $\bar{d}=- 1.75$, $\mu_d = 0$, $s_d = 7.12$, and $n = 35$. $t=\frac{-1.75 - 0}{7.12/\sqrt{35}}\approx\frac{-1.75}{7.12/5.916}\approx\frac{-1.75}{1.2035}\approx - 1.454$

Step2: Determine the degrees of freedom and find the P - value

The degrees of freedom for a paired - samples t - test is $df=n - 1=35 - 1 = 34$. Since our alternative hypothesis is $H_a:\mu_{diff}>0$, we are looking for the area to the right of the calculated t - value. Because the t - distribution is symmetric, the area to the right of $t=-1.454$ is the same as the area to the left of $t = 1.454$. Looking up $t = 1.454$ with $df = 34$ in the t - table, we find the area to the left of $t = 1.454$ is approximately $0.9225$. So the P - value (area to the right of $t=-1.454$) is $1-(1 - 0.9225)=0.9225$.

Answer:

0.9225