Round answers to 3 decimal places where possible.

Question

You are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 15 randomly selected college students.

1

5

7

1

10

0

11

5

2

10

7

9

3

5

9

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean number of days of class that college students miss is between  and  days.

c. If many groups of 15 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of missed class days and about  percent will not contain the true population mean number of missed class days. 

 

Answer 1

a. We do not know the population standard deviation so we must use a t-distribution.

b. The formula for any confidence interval for a population mean when using a t-distribution is

x-bar± t^*(s/√n) where t^* is the critical value of t with n-1 degrees of freedom.

Using your data,

x-bar = 5.667

s = 3.677

n = 15 so df = n-1 = 14

Using a t-table or an inverse t command on a calculator find t^*for a 90% CI with 14 df.

t^* = 1.761

Let's look at our formula and plug in our values.

x-bar± t^*(s/√n)

5.667± 1.761(3.677/√15)

5.667± 1.672

5.667 - 1.672 = 3.995 5.667 + 1.672 = 7.339

(3.995,7.339)

With 90% confidence the population mean number of days of class that college students miss is between 3.995  and 7.339 days.

c. Our confidence level is the percent of such intervals that would capture the true mean in repeated sampling, so the answer to the first part of the question is 90%. The second part of the question is asking what percent is left out of the first part. Since 100% - 90 % = 10%. The answer is 10%.