Question

n a study of academic procrastination, it was determined that for a random sample of 18 undergrad students at a mid-size public university, the mean time spent studying for the final exam in an intro statistics course was 7.74 hours with a standard deviation of 3.40 hours. Suppose we wish to compute a 95% confidence interval for the mean time spent studying for the intro statistics final exam by all students. The margin of error for this confidence interval would then be calculated using the critical value (c) determined in the previous question as: (Don't worry if your sample size changed- use the one given here).

Answer 1

Answer:

The 95% confidence interval for the mean time spent studying for the intro statistics final exam by all students is between 6.05 hours and 9.84 hours.

Step-by-step explanation:

We have the standard deviation for the sample, which meas 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 = 18 - 1 = 17

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.11

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.11\frac{3.4}{\sqrt{18}} = 1.69$

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 7.74 - 1.69 = 6.05 hours

The upper end of the interval is the sample mean added to M. So it is 7.74 + 1.69 = 9.84 hours.

The 95% confidence interval for the mean time spent studying for the intro statistics final exam by all students is between 6.05 hours and 9.84 hours.