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 test statistic for testing the hypotheses $h_0:mu_{diff}=0;h_a:mu_{diff}<0$?\n$t=\frac{-1.75 - 0}{sqrt{\frac{7.12}{35}}}$\n$t=\frac{0 + 1.75}{\frac{7.12}{sqrt{35}}}$\n$t=\frac{-1.75 - 0}{\frac{7.12}{sqrt{35}}}$\n$t=\frac{1.75 - 0}{\frac{7.12}{sqrt{35}}}$

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 test statistic for testing the hypotheses $h_0:mu_{diff}=0;h_a:mu_{diff}<0$?\n$t=\frac{-1.75 - 0}{sqrt{\frac{7.12}{35}}}$\n$t=\frac{0 + 1.75}{\frac{7.12}{sqrt{35}}}$\n$t=\frac{-1.75 - 0}{\frac{7.12}{sqrt{35}}}$\n$t=\frac{1.75 - 0}{\frac{7.12}{sqrt{35}}}$

Answer

Explanation:

Step1: Recall the formula for t - test statistic in paired - samples

The formula for the t - test statistic in a paired - samples t - test is $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}$, 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.

Step2: Identify the given values

We are given that $\bar{d}=- 1.75$ (the mean difference in exam scores), $\mu_d = 0$ (from the null hypothesis $H_0:\mu_{Diff}=0$), $s_d = 7.12$ (the standard deviation of the differences), and $n = 35$ (the number of pairs).

Step3: Substitute the values into the formula

Substituting the values into the formula $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}$, we get $t=\frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}$.

Answer:

$t=\frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}$ (the third option)