An organization surveyed 559 high school seniors from a certain country and foun

Question

An organization surveyed 559 high school seniors from a certain country and found that 296 believed they would not have enough money to live comfortably in college. The folks at the organization want to know if this represents sufficient evidence to conclude a majority (more than 50%) of high school seniors in the country believe they will not have enough money in college. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the =0.05 level of significance, compute the probability of making a Type II error, , if the true population proportion is 0.54. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.580.58.

Explanation / Answer

H0 : p = 0.50

Ha : p > 0.50

sample Proportion of students those who doesn't have enough money to live comfortably in college p^ = 296/559 = 0.52952

(a) Here the meaning of type II test is true proportion of students those who doesn't have enough money to live comfortably in college is higher than 0.50 but we couldn't bring evidence or we failed to reject the null hypothesis .

(b) True Population proportion p'0 = 0.54

Hypothesised proportion p0 = 0.50

so here significance level to reject null hypothesis is 0.05

standard error of the proportion se0  = sqrt[ p0 (1- p0)/ N] = sqrt [0.5 * 0.5/ 559] = 0.02115

so the acceptance region for one sided sample proportion = p^ + Z95% se0 = 0.5 + 1.645 * 0.02115 = 0.5348

so we shall not reject the null hypothesis when p < 0.5348

Now true population proportion = 0.54

standard error of the proportion se0 = sqrt [ 0.54 * 0.46/ 559] = 0.021

So Pr(Type II error) = Pr(p < 0.5348 ; 0.54 ; 0.021)

Z = (0.5348 - 0.54)/ 0.021 = -0.25

Pr(Type II error) = Pr( Z < -0.25) = 0.4013

so Pr(Type II error ) = 0.4013

Power of the test = 1 - Pr(Type II error )  = 1 - 0.4013 = 0.5987

(c) If true population proportion = 0.58

standard error of the proportion se0 = sqrt [ 0.58 * 0.42/ 559] = 0.02088

So Pr(Type II error) = Pr(p < 0.5348 ; 0.58 ; 0.02088)

Z = (0.5348 - 0.58)/ 0.02088 = -2.16

Pr(Type II error) = Pr( Z < -2.16) = 0.0154

so Pr(Type II error ) = 0.0154

Power of the test = 1 - Pr(Type II error) = 1 - 0.0154 = 0.9846

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