Respuesta :
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 [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex]. So we have T = 2.36
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.36\frac{3.4}{\sqrt{102}} = 0.79[/tex]
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.
