A random sample of size 30 from a normal population yields $\bar { X }$
= 32.8 with a population standard deviation of 4.51.Construct a 95 percent confidence interval for ?.
A) $\left[ \begin{array} { l l }
23.96 & 41.64
\end{array} \right]$
B) $[ 32.04$$ \quad$$ \quad$$33.56 ]$
C) $[ 31.45 \quad 34.15 ]$
D) $\left[ \begin{array} { l l }
31.19 & 34.41
\end{array} \right]$

Question

A random sample of size 30 from a normal population yields $\bar { X }$
= 32.8 with a population standard deviation of 4.51.Construct a 95 percent confidence interval for ?.
A) $\left[ \begin{array} { l l } 
23.96 & 41.64
\end{array} \right]$
B) $[ 32.04$$ \quad$$ \quad$$33.56 ]$
C) $[ 31.45 \quad 34.15 ]$
D) $\left[ \begin{array} { l l } 
31.19 & 34.41
\end{array} \right]$

Quizplus · Accepted Answer

Since we are constructing a confidence interval for the population mean with a known standard deviation, we can use the formula:

$$\bar{X} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$

where $\bar{X}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z_{\alpha/2}$ is the critical value from the standard normal distribution at the desired confidence level.

For a 95% confidence interval, we have $\alpha = 1 - 0.95 = 0.05$ and using a standard normal distribution table or a calculator, we can find that $z_{\alpha/2} = 1.96$.

Plugging in the values given in the problem, we get:

$$32.8 \pm 1.96 \cdot \frac{4.51}{\sqrt{30}}$$

Simplifying the expression, we get

$[31.19 \quad 34.41]$

So the best choice is D.

A Random Sample of Size 30 from a Normal Population Xˉ\bar { X }Xˉ

A Random Sample of Size 30 from a Normal Population $\bar { X }$