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1%Fri 3:35 Do Homework - Ricky Ciotti e Secure https://www.mathxl.com/Student/PlayerHomework.aspx?homeworkld-439603902&questionld; 4&flushed-true;&cld; 4602349&centerwi..; MTH 161 Online Fall 2017 Ricky Ciotti 10/20/17 3:35 PM Homework: Section 17.2 Homework Score: 0.2 of 1 pt 0%) *10.6.9 Save 40f 5 (5 complete) Score: 84% 4.2 of 5 pts Question Help * An organization surveyed 513 high school seniors from a certain country and found that 272 belileved they would not have enough money to live comfortably in college. The folks at the organization want to now if this represents suficient evidence to conclude a majority more than S0% oft hat d se on n te court 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 hypothosis at the 0.05 level of significance, compute the probability of making a Type t proportion is 0.54. What is the power of the test? Type il error. ,he true populaton at the -o 05 level of significance, compute the probability of making (c) Redo part (b) if the true population proportion is 0.58

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^ = 272/513 = 0.53

(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/ 513] = 0.022

so the acceptance region for one sided sample proportion = p^ + Z95% se0 = 0.50 + 1.645 * 0.022 = 0.5362

so we shall not reject the null hypothesis when p < 0.5362

Now true population proportion = 0.54

standard error of the proportion se0 = sqrt [ 0.54 * 0.46/ 513] = 0.022

So Pr(Type II error) = Pr(p < 0.5362 ; 0.54 ; 0.022)

Z = (0.5362 - 0.54)/ 0.022 = -0.17

Pr(Type II error) = Pr( Z < -0.17) = 0.4325

so Pr(Type II error ) = 0.4325

Power of the test = 1 - Pr(Type II error )  = 1 - 0.4325 = 0.5675

(c) If true population proportion = 0.58

standard error of the proportion se0 = sqrt [ 0.58 * 0.42/ 513] = 0.0218

So Pr(Type II error) = Pr(p < 0.5362 ; 0.58 ; 0.0218)

Z = (0.5362 - 0.58)/ 0.0218 = -2.01

Pr(Type II error) = Pr( Z < -2.01) = 0.0222

so Pr(Type II error ) = 0.0222

Power of the test = 1 - Pr(Type II error) = 1 - 0.0222 = 0.9778

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