Question

1. a teacher is interested in the effect of a study session on quiz performance. two different classes receive a pretest (before the study session) and a posttest (after the study session). thus, the teacher records the following four sets of scores:

class 1 pretest class 1 posttest
14 18
5 20
12 20
12 20
8 24
9 18
10 20
class 2 pretest class 2 posttest
20 10
2 21
12 12
8 28
12 21
4 20
12 28
a. for each of the four sets of scores above, calculate the sample mean and standard deviation.
b. in which class is the effect of the study session most obvious in the pattern of data? explain.

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.