Question

A random sample of 102 full-grown lobsters had a mean weight of 16 ounces and a standard deviation of 3.4 ounces. Construct a 98 percent confidence interval for the population mean μ. The best point estimate for a confidence interval estimating the population μ is____ounces.

Answer 1

Answer:

The best point estimate for a confidence interval estimating the population μ is 16 ounces.

The 98 percent confidence interval for the population mean μ is between 15.21 ounces and 16.79 ounces.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The best point estimate for a confidence interval estimating the population μ is

The sample mean, so 16 ounces.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 102 - 1 = 101

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 101 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$. So we have T = 2.36

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.36\frac{3.4}{\sqrt{102}} = 0.79$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 16 - 0.79 = 15.21 ounces

The upper end of the interval is the sample mean added to M. So it is 16 + 0.79 = 16.79 ounces

The 98 percent confidence interval for the population mean μ is between 15.21 ounces and 16.79 ounces.