Question

Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enrolled in. The students collect a random sample of 45 students. The mean of the sample is 12.1 units. The sample has a standard deviation of 1.5 units. What is the 95% confidence interval for the average number of units that students in their college are enrolled in

Answer 1

Answer:

The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution 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 = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2\frac{1.5}{\sqrt{45}} = 0.4$

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 12.1 - 0.4 = 11.7 units

The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.4 = 12.5 units

The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.