Question

Confidence Interval with Known $\sigma$. In Exercises 37 and 38, find the confidence interval using the known value of $\sigma$.

Birth Weights of Girls Construct the confidence interval for Exercise 9 “Birth Weights of Girls,” assuming that $\sigma$is known to be 7.1 hg.

Short answer

The 95% confidence interval for the estimate mean is$29.4 \text{hg}<\mu <31.4 \text{hg}$.

Step-by-step solution

Given information

Refer to Exercise 9 for the summary statistics for randomly selected weights of newborn girls.

Here,

$n = 205,\overline{x} =30.4 \text{hg}$.

The 95% confidence level with the known value of $\sigma =7.1 \text{hg}$.

Describe confidence interval

A confidence interval is an estimate of the interval that may contain the true value of a population parameter. It is also known as an interval estimate.

The general formula for the confidence interval estimate of mean for the known$\sigma$ is as follows.

$\text{Confidence} \text{interval}=\left(\overline{x}-\text{E},\overline{x}+\text{E}\right) ...\left(1\right)$

Here, E is the margin of error, which is calculated as follows.

$\text{E}=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{\text{n}}}$

Find the appropriate distribution

For a normally distributed population with randomly selected observations, the following are true.

If $\sigma$is known, the normal distribution is suitable to find the confidence interval.

If $\sigma$is unknown, the student’s t-distribution is suitable to find the confidence interval.

In this case, $\sigma$is known, and $\text{n}=205$, which is greater than 30.

Thus, normal distribution applies.

Find the critical value zα2

$z_{\frac{\alpha }{2}}$is a z score that separates an area of $\frac{\alpha }{2}$in the right tail of the standard normal distribution.

The confidence level 95% corresponds to $\alpha =0.05 \text{and} \frac{\alpha }{2}=0.025$.

The value$z_{\frac{\alpha }{2}}$has the cumulative area $1-\frac{\alpha }{2}$to its left. .

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{\alpha }{2}}\right)=1-\frac{\alpha }{2} \\ =0.975 \end{gathered}$$

From the standard normal table, the area of 0.975 is observed corresponding to the row value 1.9 and column value 0.06, which implies that role="math" localid="1648033895973" $z_{\frac{\alpha }{2}}$is 1.96.

Find the margin of error

The margin error is calculated as follows.

$$\begin{gathered} \text{E}=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{\text{n}}} \\ =1.96\times \frac{7.1}{\sqrt{205}} \\ =0.9719 \end{gathered}$$

Find the confidence interval

The confidence interval is obtained by substituting the value of the margin of error in equation (1), as follows.

$$\begin{gathered} \text{Confidence} \text{interval}=\left(\overline{x}-E,\overline{x}+E\right) \\ =\left(30.4-0.9719 , 30.4+0.9719\right) \\ =\left(29.4281 , 31.3719\right) \end{gathered}$$

Thus, the 95% confidence interval for the estimate mean is $29.4 \text{hg}<\mu <31.4 \text{hg}$.