Question

The heights of seasonal pine saplings are normally distributed with an unknown population mean and standard deviation. A random sample of 20 saplings is taken and results in a sample mean of 310 millimeters and sample standard deviation of 20 millimeters. (a) The EBM, margin of error, for a 98% confidence interval estimate for the population mean using the Student's t-distribution is 11.35. (b) Find a 98% confidence interval estimate for the population mean using the Student's t-distribution. Round the final answers to two decimal places.

Answer 1

a) The margin of error for the interval is of 11.35.

b) The 98% confidence interval estimate for the population mean using the Student's t-distribution is between 298.65 and 321.35 millimeters.

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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 = 20 - 1 = 19

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.98)/(2) = 0.99`. So we have T = 2.539

The margin of error is:

`M = T(s)/(√(n)) = 2.539(20)/(√(20)) = 11.35`

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

Thus, the margin of error for the interval is of 11.35.

The lower end of the interval is the sample mean subtracted by M. So it is 310 - 11.35 = 298.65 millimeters.

The upper end of the interval is the sample mean added to M. So it is 310 + 11.35 = 321.35 millimeters.

The 98% confidence interval estimate for the population mean using the Student's t-distribution is between 298.65 and 321.35 millimeters.