Question

150 subjects in a psychological study had a mean score of 36 on a test instrument designed to measure anger. The standard deviation of the 150 scores was 12. Find a 98% confidence interval for the mean anger score of the populations from which the sample was selected.

Select one: a. (34.1, 37.9) b. (33.5, 38.5) c. (33.7, 38.3) d. None of other answer is necessary true.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 36

Population standard deviation =    = 12

Sample size = n =150

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2 * ( /n)

= 2.326 * ( 12 /  150 )

E= 2.2790
At 98% confidence interval estimate of the population mean
is,

- E < < + E

36 -2.2790 <   <36 +2.2790

33.7 <   < 38.3

(33.7, 38.3)