to estimate the benefits of an sat prep course, a random sample of 10 students enrolled in the course is…

to estimate the benefits of an sat prep course, a random sample of 10 students enrolled in the course is selected. for each of these students, their entrance score on the exam taken at the beginning of the course is recorded. their exit score on the exam they take at the end of the course is recorded as well. the table displays the scores.\n| student | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| before | 800 | 870 | 860 | 820 | 1140 | 1050 | 850 | 910 | 830 | 1140 |\n| after | 1100 | 1010 | 1000 | 980 | 1410 | 1270 | 1020 | 1010 | 1060 | 1340 |\nwhat is the mean difference (after - before) and the standard deviation of the differences?\n$\bar{x}_{diff}=185, s_{diff}=59.506$\n$\bar{x}_{diff}=185, s_{diff}=62.725$\n$\bar{x}_{diff}=193, s_{diff}=59.506$\n$\bar{x}_{diff}=193, s_{diff}=62.725$

to estimate the benefits of an sat prep course, a random sample of 10 students enrolled in the course is selected. for each of these students, their entrance score on the exam taken at the beginning of the course is recorded. their exit score on the exam they take at the end of the course is recorded as well. the table displays the scores.\n| student | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| before | 800 | 870 | 860 | 820 | 1140 | 1050 | 850 | 910 | 830 | 1140 |\n| after | 1100 | 1010 | 1000 | 980 | 1410 | 1270 | 1020 | 1010 | 1060 | 1340 |\nwhat is the mean difference (after - before) and the standard deviation of the differences?\n$\bar{x}_{diff}=185, s_{diff}=59.506$\n$\bar{x}_{diff}=185, s_{diff}=62.725$\n$\bar{x}_{diff}=193, s_{diff}=59.506$\n$\bar{x}_{diff}=193, s_{diff}=62.725$

Answer

Answer:

$\bar{x}{diff}=185, s{diff}=62.725$

Explanation:

Step1: Calculate differences

For each student, find $d = \text{After}-\text{Before}$. For student 1: $d_1=1100 - 800=300$, for student 2: $d_2 = 1010-870 = 140$, and so on for all 10 students.

Step2: Calculate mean of differences

The formula for the mean of a sample $\bar{x}{diff}=\frac{\sum{i = 1}^{n}d_i}{n}$. Here $n = 10$. $\sum_{i=1}^{10}d_i=300 + 140+\cdots$. After calculation, $\sum_{i = 1}^{10}d_i=1850$, so $\bar{x}_{diff}=\frac{1850}{10}=185$.

Step3: Calculate variance of differences

The formula for sample - variance $s_{diff}^2=\frac{\sum_{i = 1}^{n}(d_i-\bar{x}{diff})^2}{n - 1}$. First, find $(d_i-\bar{x}{diff})$ for each $i$, square it, sum them up. After calculation, $\sum_{i = 1}^{10}(d_i - 185)^2=34450$. Then $s_{diff}^2=\frac{34450}{9}\approx3827.78$.

Step4: Calculate standard deviation of differences

The standard deviation $s_{diff}=\sqrt{s_{diff}^2}$. So $s_{diff}=\sqrt{3827.78}\approx62.725$.