Question

A sample of 10 students record their scores on the final exam for their statistics class. The mean of the sample is 81 with sample standard deviation 7 points. Analysis of the 10 sample values indicated that the population is approximately normal. We wish to find the 95% confidence interval for the population mean test scores.

1. What is the confidence level, c?

2. Which of the following is correct?

a. To find the confidence interval, a z-critical value should be used because the population standard deviation is known.

b. To find the confidence interval, a t-critical value should be used because the population standard deviation is known.

c. To find the confidence interval, a z-critical value should be used because the population standard deviation is not known.

d. To find the confidence interval, a t-critical value should be used because the population standard deviation is not known.

3. What are the degrees of freedom?

4. Find t_c (to three decimal places).

5. Find the error, E, of the confidence interval (round your answer to 3 decimal places).

6. Which of the following gives the 95% confidence interval for the population test score?

a. (85, 86.007)

b. (80, 90)

c. (75.993, 86.007)

d. (75.993, 85)

e. (74.745, 87.334)

f. (73.485, 87.341)

Answer 1

1) C=0.95

2) d. To find the confidence interval, a t-critical value should be used because the population standard deviation is not known.

3) degree of freedom=   DF=n-1=   9

4) 't value='   tα/2=   2.262   [Excel formula =t.inv(α/2,df) ]

5) Standard Error , SE = s/√n =   7/√10=   2.2136          
margin of error , E=t*SE =   2.2622   *   2.2136   =   5.007

6) confidence interval is                   
Interval Lower Limit = x̅ - E =    81.00   -   5.0075   =   75.9925
Interval Upper Limit = x̅ + E =    81.00   -   5.0075   =   86.0075
95%   confidence interval is (   75.993   < µ <   86.007   ) (option C)