Question

people are given a standardized test, the scores result in a sample mean of 73 and a sample standard deviation of 12

a) what is the 99% confidence interval around the mean

b) the 95% confidence interval

c) possible value for u/(mu) that would be rejected at the .05 level but accepted at the .01 level

n=25

Answer 1

Given that n = 25, xbar = 73 and sigma = 12 we can write the confidence interval as:

xbar +- z(alpha/2) * sqrt(sigma^2/n)

a) For 99% confidence interval the value of alpha is 0.01. Hence, using the formula the 99% confidence interval is given as:

(66.82, 79.18)

b) The 95% confidence interval is given asL

(68.3, 77.7)

c) The possible values of mu that would be rejected at 5% level of significance but accepted at 1% level of significance are:

(66.82, 68.3) U (77.7, 79.18)

The union of the above two sets.