Question

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of 1.2 days. Construct a 95% confidence interval for the population's mean length of vacation. ( , ) Round your answer to two decimal digits Construct a 92% confidence interval for the population's mean length of vacation. ( , ) Round your answer to two decimal digits

Answer 1

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations 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 = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

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 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).