Question

A genetic marker is expected to be present in approximately 1/3 of the population. A simple random sample of 70 individuals is tested, and 20 turn out to have the marker. Do a test for the Null Hypothesis p = 1/3

H0 : p = 1/3

Ha : p does not = 1/3

. Determine the “acceptance region”. Find the probability of the sample mean falling into this region if the true mean happens to be 0.4, 0.5, 0.6. This probability is called the “Error of Type II”, and is the probability of not rejecting the Null Hypothesis, when you should (it clearly varies according to what the true value of p is)

Answer 1

Standard error of sample proportion, SE = = 0.056343617

Assuming the significance level of 0.05, z value = 1.96

Critical value to reject H0 are, 1/3 - 1.96 * 0.056343617 and  1/3 + 1.96 * 0.056343617

= 0.2229 and 0.4438

That is, we reject H0 is the sample proportion, < 0.2229 or > 0.4438

Acceptance region is,

0.2229 < < 0.4438

When true mean is 0.4,

Standard error of sample proportion, SE = = 0.0585540044

Probability of the sample mean falling into this region if the true mean happens to be 0.4

= P(0.2229 < < 0.4438)

= P( < 0.4438) - P( < 0.2229)

= P[Z < (0.4438 - 0.4)/0.0585540044] - P[Z < (0.2229 - 0.4)/0.0585540044]

= P[Z < 0.75] - P[Z < -3.02]

= 0.7734 - 0.0013

= 0.7721

When true mean is 0.5,

Standard error of sample proportion, SE = = 0.0597614305

Probability of the sample mean falling into this region if the true mean happens to be 0.5

= P(0.2229 < < 0.4438)

= P( < 0.4438) - P( < 0.2229)

= P[Z < (0.4438 - 0.5)/0.0597614305] - P[Z < (0.2229 - 0.5)/0.0597614305]

= P[Z < -0.94] - P[Z < -4.64]

= 0.1736 - 0

= 0.1736

When true mean is 0.6,

Standard error of sample proportion, SE = = 0.0585540044

Probability of the sample mean falling into this region if the true mean happens to be 0.6

= P(0.2229 < < 0.4438)

= P( < 0.4438) - P( < 0.2229)

= P[Z < (0.4438 - 0.6)/0.0585540044] - P[Z < (0.2229 - 0.6)/0.0585540044]

= P[Z < -2.67] - P[Z < -6.44]

= 0.0038 - 0

= 0.0038