Save \& Exit Certify Lesson: 8.3 Estimating Population Means (... Question 8 of 10, Step 4 of 4 A certain test preparation course is designed to help students improve their scores on the GRE exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 7 students' scores on the exam after completing the course: $$ 24,6,21,8,16,10,18 $$ Using these data, construct a $$ 80 \% $$ confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 4 of 4 : Construct the $$ 80 \% $$ confidence interval. Round your answer to one decimal place.

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We are given the net change scores for 7 students:

$$
24,\;6,\;21,\;8,\;16,\;10,\;18
$$

**Step 1. Calculate the sample mean.**

The sample mean is given by

$$
\bar{x}=\frac{24+6+21+8+16+10+18}{7}=\frac{103}{7}\approx 14.7.
$$

**Step 2. Calculate the sample standard deviation.**

First, compute the deviations from the mean:

$$
\begin{array}{ll}
24 - 14.7 \approx 9.3, & 6 - 14.7 \approx -8.7, \
21 - 14.7 \approx 6.3, & 8 - 14.7 \approx -6.7, \
16 - 14.7 \approx 1.3, & 10 - 14.7 \approx -4.7, \
18 - 14.7 \approx 3.3. &
\end{array}
$$

Now, square these differences:

$$
\begin{array}{l}
(9.3)^2 \approx 86.5, \
(-8.7)^2 \approx 75.7, \
(6.3)^2 \approx 39.7, \
(-6.7)^2 \approx 44.9, \
(1.3)^2 \approx 1.7, \
(-4.7)^2 \approx 22.1, \
(3.3)^2 \approx 10.9.
\end{array}
$$

Sum the squared differences:

$$
86.5+75.7+39.7+44.9+1.7+22.1+10.9\approx 281.5.
$$

The sample variance is

$$
s^2=\frac{281.5}{7-1}=\frac{281.5}{6}\approx 46.9,
$$

so the sample standard deviation is

$$
s\approx \sqrt{46.9}\approx 6.8.
$$

**Step 3. Compute the standard error of the mean.**

$$
SE=\frac{s}{\sqrt{n}}=\frac{6.8}{\sqrt{7}}\approx \frac{6.8}{2.65}\approx 2.6.
$$

**Step 4. Determine the appropriate $$ t $$-value.**

Since we are constructing an $$80\%$$ confidence interval with $$ n=7 $$ (degrees of freedom $$ df = 6 $$) and $$\alpha = 0.20$$ (with $$\alpha/2=0.10$$ in each tail), we use the $$t$$-distribution. For $$ df=6 $$, the critical $$ t $$-value $$ t_{0.10,6} $$ is approximately

$$
t^*\approx 1.44.
$$

**Step 5. Construct the confidence interval.**

The confidence interval is given by

$$
\bar{x}\pm t^*\times SE.
$$

Substitute the values:

$$
\text{Margin of error} = 1.44\times 2.6\approx 3.7.
$$

Thus, the $$80\%$$ confidence interval is

$$
14.7\pm 3.7.
$$

Calculating the endpoints:

$$
\begin{aligned}
\text{Lower limit}&: 14.7-3.7\approx 11.0,\[5mm]
\text{Upper limit}&: 14.7+3.7\approx 18.4.
\end{aligned}
$$

**Final Answer:**

$$
[11.0,\;18.4]
$$
The 80% confidence interval for the average net change in a student's score after completing the course is [11.0, 18.4].

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