Question

Q15 Determine the positive critical value of z that would be used in developing a confidence interval of the difference between two population proportions, for large independent samples where the desired level of significance (alpha) is 0.06. Use Excel's NORMSINV function, check it against the values that would be nearest in the standard normal distribution table, and enter your response to two decimal places (i.e. 2.12).

Answer 1

As we are trying to find a confidence interval of the difference between two population proportions, for large independent samples where the desired level of significance (alpha) is 0.06, therefore we need to find K such that:

P( -K < Z < K) = 1 - 0.06

P( -K < Z < K) = 0.94

Due to symmetry, we get here:

P(Z < K) = 0.94 + 0.06/2 = 0.97

Therefore Using the EXCEL function, we have:
=NORMINV(0.97,0,1)

The required value of critical value here is 1.8808

Now from the standard normal tables, we have:

P(Z < 1.88) = 0.9699
P(Z < 1.89) = 0.9706

Therefore 1.88 is the required positive critical value here.