91影视

Refer to Exercise 10 for the summary statistics for randomly selected weights of boys as .

The 95% confidence level with the known value of $n鈥�=鈥�195,\stackrel{炉}{x}鈥�=32.7鈥�\text{hg}$.

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 known$蟽$ is as follows.

$\text{Confidence}鈥�鈥�\text{interval}=\left(\stackrel{炉}{x}-\text{E},\stackrel{炉}{x}+\text{E}\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

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

$\text{E}=z_{\frac{伪}{2}}脳\frac{蟽}{\sqrt{\text{n}}}$

For a normally distributed population and randomly selected samples, the following are true.

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

If $蟽$is unknown, the student鈥檚 t-distribution is suitable to find the confidence interval.

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

Thus, normal distribution applies.

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

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

The value$z_{\frac{伪}{2}}$hasthe cumulative area $1-\frac{伪}{2}$to its left. .

Mathematically,

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

From the standard normal table, the area of 0.975 is observed as the corresponding intersection of the row value 1.9 and column value 0.06, which implies that role="math" localid="1648035872680" $z_{\frac{伪}{2}}$is 1.96.

The margin of error is calculated as follows.

$$\begin{gathered} \text{E}=z_{\frac{伪}{2}}脳\frac{蟽}{\sqrt{\text{n}}} \\ =1.96脳\frac{6.6}{\sqrt{195}} \\ =0.9264 \end{gathered}$$

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

$$\begin{gathered} \text{Confidence}鈥�鈥�\text{Interval}=\left(\stackrel{炉}{x}-\text{E},\stackrel{炉}{x}+\text{E}\right) \\ =\left(32.7-0.9264鈥�鈥�,鈥�鈥�32.7+0.9264\right) \\ =\left(31.7736鈥�,鈥�鈥�33.6264\right) \end{gathered}$$

Thus, the 95% confidence interval for the estimate mean is$31.8鈥�\text{hg}<渭<33.6鈥�\text{hg}$.