Question

Please answer questions 4 & 5

Information The following applies to Questions 1 to 5: In a random sample of 150 Manitoba high school students,it is found that 66 of them have enrolled in university Question 1 (1 point) what is the margin of error for a 97% confidence interval for the true proportion of Manitoba high school students who enroll in university? Report your answer rounded to 3 decimal places. Your Answer: 0.144 Answer

Answer 1

QUestion4

True proportion = 0.37

Test is conducted at the significance level 0.01

so we will reject the null hypothesis when p^ < 0.50 - Z_0.01 * sqrt(0.5 * 0.5/150)

p^ < 0.50 - 2.326 * 0.0408

p^ < 0.4050

Here standard error of proportion when true proportion is 0.37 = sqrt (0.37 * 0.63/150) = 0.0394

Power of the test = Pr(p^ < 0.405) = NORM (p^ < 0.405 ; 0.37 ; 0.0394)

Z = (0.405 - 0.37)/0.0394 = 0.8254

Pr(p^ < 0.405) = Pr(Z < 0.8254) = 0.7954

Question 5

Null hypothesis will be rejected if p^ <= 0.4453

so Here the Z - value

Z = (0.4453 - 0.5)/ sqrt (0.5 * 0.5/150)

Z = -1.3399

so Significance level = Pr(Z < -1.3399) = 0.09

so here the significance level would be 0.09.