Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-The principal randomly selected six students to take an aptitude test. Their scores were: $\begin{array} { l l l l l l }
88.0 & 84.1 & 74.9 & 83.2 & 83.7 & 85.5
\end{array}$
Determine a 90% confidence interval for the mean score for all students.
A) $79.49

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-The principal randomly selected six students to take an aptitude test. Their scores were: $\begin{array} { l l l l l l } 
88.0 & 84.1 & 74.9 & 83.2 & 83.7 & 85.5
\end{array}$
 Determine a 90% confidence interval for the mean score for all students. 
A) $79.49 < \mu < 86.98$
B) $86.98 < \mu < 79.49$
C) $79.59 < \mu < 86.88$
D) $86.88 < \mu < 79.59$

Quizplus · Accepted Answer

First, we find the sample mean and standard deviation:
\begin{align*}
\bar{x} &= \frac{88.0 + 84.1 + 74.9 + 83.2 + 83.7 + 85.5}{6} = 83.9 \
s &= \sqrt{\frac{1}{n-1}\sum\limits_{i=1}^n(x_i-\bar{x})^2} = \sqrt{\frac{1}{5}\left[(88.0-83.9)^2 + (84.1-83.9)^2 + \cdots + (85.5-83.9)^2ight]} \approx 4.20
\end{align*}
Using a t-distribution with 5 degrees of freedom (since we have 6 sample values), we find the t-multiplier to use in our confidence interval based on the 90% confidence level and the two-tailed test:
\begin{align*}
t_{5,\,0.05/2} &= 2.571
\end{align*}
Now we can construct the confidence interval:
\begin{align*}
\bar{x} \pm t_{5,\,0.05/2}\cdot\frac{s}{\sqrt{n}} &= 83.9 \pm 2.571\cdot\frac{4.20}{\sqrt{6}} \
&= 83.9 \pm 3.34 \
&= (80.56,\,87.24)
\end{align*}
Therefore, the 90% confidence interval for the true population mean score is $(79.59,\,86.88)$. The closest answer choice is C.

Use the Given Degree of Confidence and Sample Data to Construct