A laboratory tested a sample of 82 duck eggs and found that the mean amount of cholesterol was 228 milligrams. Suppose the population standard deviation of the amount of cholesterol is known to be 19.0 milligrams.

Construct a 99% confidence interval for the true population mean cholesterol content of all such eggs and use one decimal. Write your interval in the form (lower limit, upper limit)

Solution

Given that sample size $n = 82$, sample mean $\overline{X}= 228$ and population standard deviation $\sigma = 19$.

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

Step 2 Given information

Given that sample size $n =82$, sample mean $\overline{X}=228$ and population standard deviation is $\sigma = 19$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$
where $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

z-critical values
z-critical values

Thus $Z_{\alpha/2} = Z_{0.005} = 2.58$.

Step 5 Compute the margin of error

The margin of error for mean is

$$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 2.58 \frac{19}{\sqrt{82}} \\ & = 5.405. \end{aligned} $$

Step 6 Determine the confidence interval

$99$% confidence interval estimate for population mean is

$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 228 - 5.405 & \leq \mu \leq 228 + 5.405\\ 222.595 & \leq \mu \leq 233.405. \end{aligned} $$

Thus, $99$% confidence interval for the true population mean cholesterol content of all such eggs is $(222.595,233.405)$.

Further Reading