Question

Suppose that the number of square feet per house are normally distributed with an unknown mean and standard deviation. A random sample of 22 houses is taken and gives a sample mean of 1500 square feet and a sample standard deviation of 151 square feet. 1. The EBM, margin of error, for a 95% confidence interval estimate for the population mean using the Student's t. distribution is 66.96.2. Find a 95% confidence interval estimate for the population mean using the Student's t-distribution.

Answer 1

Answer:

1. The margin of error is of 66.96 square feet.

2. The 95% confidence interval estimate for the population mean using the Student's t-distribution is between 1433.04 square feet and 1566.96 square feet

Step-by-step explanation:

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 = 22 - 1 = 21

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.08

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.08*\frac{151}{\sqrt{22}} = 66.96$

In which s is the standard deviation of the sample.

The margin of error is of 66.96 square feet.

The lower end of the interval is the sample mean subtracted by M. So it is 1500 - 66.96 = 1433.04 square feet

The upper end of the interval is the sample mean added to M. So it is 1500 + 314 = 1566.96 square feet

The 95% confidence interval estimate for the population mean using the Student's t-distribution is between 1433.04 square feet and 1566.96 square feet