homework: section 8.2
score: 1.5/12
question 2
use the following sample to estimate a population mean μ.
find the 99% confidence interval about the population mean. enter your answer as a tri - ling inequality accurate to two decimal place.
submit question

Question

Accepted Answer

# Explanation:
## Step1: Calculate the sample mean $\bar{x}$
First, count the number of data points $n$. There are $41$ data points.
Sum up all the data values:
$$
\begin{align*}
&\sum x=24.8 + 26.7+24.5 + 15.8+25.4+21.2+11.8+16+18.8+17.4+26.9+18+26.6+29.2+17.1+24.1+18+14.6+23.3+26.4+18.9+14.3+16.7+22.3+19.7+14.2+21.6+21.5+15.4+20.4+16.6+23.3+28.7+21+22.9+25.3+16.4+16.3+27.6+22.4+14.5+15.2+15.9\
&=839.9
\end{align*}
$$
The sample mean $\bar{x}=\frac{\sum x}{n}=\frac{839.9}{41}\approx20.49$

## Step2: Calculate the sample standard deviation $s$
$$
\begin{align*}
s&=\sqrt{\frac{\sum(x - \bar{x})^{2}}{n- 1}}\
\end{align*}
$$
$$
\begin{align*}
\sum(x-\bar{x})^{2}&=(24.8 - 20.49)^{2}+(26.7-20.49)^{2}+\cdots+(15.9 - 20.49)^{2}\
\end{align*}
$$
After calculation, $\sum(x-\bar{x})^{2}\approx1077.91$
$s=\sqrt{\frac{1077.91}{40}}\approx5.20$

## Step3: Determine the critical value $t$
For a $99\%$ confidence interval and $n - 1=40$ degrees of freedom, using a $t$-distribution table or calculator, the critical value $t_{\alpha/2}$ with $\alpha=1 - 0.99 = 0.01$ and $\frac{\alpha}{2}=0.005$, $t_{0.005,40}\approx2.704$

## Step4: Calculate the margin of error $E$
The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$
$E = 2.704\times\frac{5.20}{\sqrt{41}}\approx2.704\times0.81\approx2.19$

## Step5: Calculate the confidence interval
The confidence interval is $\bar{x}-E<\mu<\bar{x}+E$
$20.49-2.19<\mu<20.49 + 2.19$

# Answer:
$18.30<\mu<22.68$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the sample mean \(\bar{x}\)

Step2: Calculate the sample standard deviation \(s\)

Step3: Determine the critical value \(t\)

Step4: Calculate the margin of error \(E\)

Step5: Calculate the confidence interval

Answer: