Question

please right the answer that i can understand please

In December 2004, 39% of students in high school were satisfied with the lunches supplied through the school. In May 2010, an organization conducted a poll of 851 students in high school and asked if they were satisfied with the lunches supplied through the school. Of the 851 surveyed, 298 indicated they were satisfied. Does this suggest the proportion of students satisfied with the quality of lunches has decreased? (a) What does it mean to make a Type Il error for this test? (b) of the researcher decides to test this hypothesis at the c = 0.10 level of significance, compute the probability of making a Type Il error, B. if the true population proportion is 0.35. What is the power of the test? (c) Redo part(b) if the true population proportion is 0.38. (a) What does it mean to make a Type Il error for this test? Choose the correct answer below. O A. Ho is rejected and the true proportion of high school students who are not satisfied with the quality of lunches is less than 0.39. O B. Ho is not rejected and the true proportion of high school students who are not satisfied with the quality of lunches is equal to 0.39. O C. Ho is not rejected and the true proportion of high school students who are not satisfied with the quality of lunches is less than 0.39. (b) of the researcher decides to test this hypothesis at the q=0.10 level of significance, compute the probability of making a Type Il error, B, if the true population proportion is 0.35. What is the power of the test? B = 0.05 Power 0.3334 (Type Integers or decimals rounded to four decimal places as needed.) (c) Redo part (b) the true population proportion is 0.38. B = 0.001 Power = 2.3638

Answer 1

(a) Type II error means fail to reject null hypothesis when null hypothesis is false. So Option C.

Your answer is correct.

(b)

V We reject Ho if P -0.39 = < -20.10 = -1.2816 0.39+(1-0.39) 851 0.39* (1 -0.39) or, P < 0.39 - 1.2816* V- -=0.3686. V 851 Probability of Type 11 error when p=0.35 = P > 0.3686|p=0.35) = 1-0 ((0.3686 -0.35)/0.35*(1 -0.35)/851) = 0.1276 Power at p=0.35 = 1 - Probability of Type II error when p=0.35 = 1-0.1276 = 0.8724

(c) We reject H, if P - 0.39 = < -20.10 = -1.2816 0.39+(1-0.39) 851 851 0.39 * (1 -0.39) or, p < 0.39 – 1.2816* V = 0.3686. Probability of Type II error when p=0.38 = PP > 0.3686|p=0.38) = 1-0 (0.3686 -0.38)/V0.38 *(1 -0.38)/851) = 0.7534 Power at p=0.38 = 1 - Probability of Type II error when p=0.38 =1-0.7534 = 0.2466