38. a teacher is interested in the effect of a study session on quiz performance. two different classes…

Explanation:

Step1: Calculate Class 1 Pretest mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For Class 1 Pretest scores ($14,5,12,12,8,9,10$), $n = 7$, $\sum_{i=1}^{7}x_{i}=14 + 5+12+12+8+9+10=70$. So, $\bar{x}{1p}=\frac{70}{7}=10$. The formula for the sample - standard deviation $s=\sqrt{\frac{\sum{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. $(14 - 10)^{2}=16$, $(5 - 10)^{2}=25$, $(12 - 10)^{2}=4$, $(12 - 10)^{2}=4$, $(8 - 10)^{2}=4$, $(9 - 10)^{2}=1$, $(10 - 10)^{2}=0$. $\sum_{i = 1}^{7}(x_{i}-\bar{x}{1p})^{2}=16 + 25+4+4+4+1+0=54$. $s{1p}=\sqrt{\frac{54}{6}}=\sqrt{9}=3$.

Step2: Calculate Class 1 Posttest mean

For Class 1 Posttest scores ($18,20,20,20,24,18,20$), $n = 7$, $\sum_{i=1}^{7}x_{i}=18+20+20+20+24+18+20 = 140$. So, $\bar{x}{1s}=\frac{140}{7}=20$. $(18 - 20)^{2}=4$, $(20 - 20)^{2}=0$, $(20 - 20)^{2}=0$, $(20 - 20)^{2}=0$, $(24 - 20)^{2}=16$, $(18 - 20)^{2}=4$, $(20 - 20)^{2}=0$. $\sum{i = 1}^{7}(x_{i}-\bar{x}{1s})^{2}=4+0+0+0+16+4+0=24$. $s{1s}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$.

Step3: Calculate Class 2 Pretest mean

For Class 2 Pretest scores ($20,2,12,8,12,4,12$), $n = 7$, $\sum_{i=1}^{7}x_{i}=20+2+12+8+12+4+12=60$. So, $\bar{x}{2p}=\frac{60}{7}\approx8.57$. $(20 - 8.57)^{2}\approx131.65$, $(2 - 8.57)^{2}\approx43.16$, $(12 - 8.57)^{2}\approx11.76$, $(8 - 8.57)^{2}\approx0.32$, $(12 - 8.57)^{2}\approx11.76$, $(4 - 8.57)^{2}\approx20.88$, $(12 - 8.57)^{2}\approx11.76$. $\sum{i = 1}^{7}(x_{i}-\bar{x}{2p})^{2}\approx131.65+43.16+11.76+0.32+11.76+20.88+11.76 = 231.39$. $s{2p}=\sqrt{\frac{231.39}{6}}\approx6.24$.

Step4: Calculate Class 2 Posttest mean

For Class 2 Posttest scores ($10,21,12,28,21,20,28$), $n = 7$, $\sum_{i=1}^{7}x_{i}=10+21+12+28+21+20+28=140$. So, $\bar{x}{2s}=\frac{140}{7}=20$. $(10 - 20)^{2}=100$, $(21 - 20)^{2}=1$, $(12 - 20)^{2}=64$, $(28 - 20)^{2}=64$, $(21 - 20)^{2}=1$, $(20 - 20)^{2}=0$, $(28 - 20)^{2}=64$. $\sum{i = 1}^{7}(x_{i}-\bar{x}{2s})^{2}=100+1+64+64+1+0+64=294$. $s{2s}=\sqrt{\frac{294}{6}}=7$.

Step5: Analyze the effect of the study session

For Class 1, the mean increased from $10$ to $20$. For Class 2, the mean increased from approximately $8.57$ to $20$. However, looking at the spread of scores, in Class 1, the pre - test scores were more spread out ($s_{1p}=3$) and post - test scores were less spread out ($s_{1s}=2$), while in Class 2, the pre - test ($s_{2p}\approx6.24$) and post - test ($s_{2s}=7$) standard deviations are relatively close. The increase in mean and the reduction in variability in Class 1 make the effect of the study session more obvious.

Answer:

a. Class 1 Pretest: Mean = 10, Standard deviation = 3 Class 1 Posttest: Mean = 20, Standard deviation = 2 Class 2 Pretest: Mean $\approx8.57$, Standard deviation $\approx6.24$ Class 2 Posttest: Mean = 20, Standard deviation = 7 b. Class 1. The mean increased from 10 to 20 and the standard deviation decreased from 3 to 2, showing a more consistent improvement in scores after the study session compared to Class 2 where the standard deviation did not change as significantly while the mean also increased.