Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.
-A random sample of 187 full-grown lobsters had a mean weight of 19 ounces and a standard deviation of 3.3 ounces. Construct a 98% confidence interval for the population mean $\mu _ {. }$
A) $18 \mathrm { oz }

Question

Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.
-A random sample of 187 full-grown lobsters had a mean weight of 19 ounces and a standard deviation of 3.3 ounces. Construct a 98% confidence interval for the population mean $\mu _ {. }$ 
A) $18 \mathrm { oz } < \mu < 21 \mathrm { oz }$
B) $19 \mathrm { oz } < \mu < 21 \mathrm { oz }$
C) $17 \mathrm { oz } < \mu < 19 \mathrm { oz }$
D) $18 \mathrm { oz } < \mu < 20 \mathrm { oz }$

Quizplus · Accepted Answer

We will 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, $\alpha$ is the significance level, and $z_{\alpha/2}$ is the critical value for the standard normal distribution at $\alpha/2$ level of significance.

Substituting the given values, we get
$$19 \pm z_{0.01/2} \frac{3.3}{\sqrt{187}}$$
The critical value $z_{0.005}$ can be found using a standard normal distribution table or calculator. It is approximately 2.576.

Substituting this value, we have:
$$19 \pm 2.576 	imes \frac{3.3}{\sqrt{187}}$$
Simplifying this expression gives:
$$19 \pm 0.98$$
Rounding to one decimal place (the same number of decimal places as the sample mean), we get:
$$18.0 < \mu < 20.0$$
So the answer is choice D.

Use the Confidence Level and Sample Data to Find a Confidence