Question

Suppose you want to test the following hypotheses: H0: p ≥ 0.4 vs. H1: p < 0.4. A random sample of 1000 observations was taken from the population. Answer the following questions and show your Excel calculation for each question clearly: (a) Let p ̂ be the sample proportion. What is the standard error of sample proportion (i.e., σ_p ̂ ) if H0 is true? (b) If the sample proportion obtained were 0.38 (i.e., p ̂=0.38), what is its p-value? (c) If your decision rule is to “reject H0 if p ̂ < 0.375 and do not reject otherwise”, what is the probability of Type I error? (d) Suppose that the population proportion is known to be 0.37 (i.e., H0 is false). What is the probability of Type II error if the decision rule given in (c) is used? (e) What is the decision rule if you were to test the above hypothesis at α = 5% significance level? Specify the cutoff value and the decision rule for accepting/rejecting H0. (f) Instead of doing a right-tailed test specified above, suppose that you want to do a two-tailed test (i.e., H0: p = 0.4 vs. H1: p ≠ 0.4) at α = 1% significance level. Specify the decision rule that can be used to reject or accept H0 (i.e., the cutoff values for accepting/rejecting H0).

Answer 1

H₀: p ≥ 0.4 vs. H₁: p < 0.4.

Sample size = n = 1000

(a) Standard error of sample proportion = = sqrt [0.4 * 0.6/1000] = 0.0155

(b) sample proportion p^ = 0.38

p - value= NORMDIST(p^ < 0.38 ; 0.4 ; 0.0155)

z = (0.38 - 0.4)/0.0155 = -1.291

p - value = P(Z < 1.291) = 0.0985

(c) Here decision rule is  to “reject H0 if p ̂ < 0.375 and do not reject otherwise”

here  Type I error occurs when the null hypothesis is correct and we reject the null hypothesis.

so here type I error = P(p^ < 0.375)

Z = (0.375 - 0.4)/0.0155 = -1.6137

type I error = P(p^ < 0.375) = P(Z < -1.6137) = 0.0533

(d) Here true proportion = p = 0.37

standard error of proportion = sqrt [0.37 * 0.63/1000] = 0.0153

so type II error occurs when the sample proportion p^ > 0.375

P(p^ > 0.375) = 1 - P(p^ < 0.375) = 1 - P(p^ < 0.375 ; 0.37 ; 0.0153)

z = (0.375 - 0.37)/0.0153 = 0.3275

Type II error = 1 - P(Z < 0.3275) = 1 - 0.6284 = 0.3716

(e) Here significance level = α = 5%

Critical test statistic = NORMSINV(0.05) = -1.645

Curoff value = p - Z_α = 0.4 - 1.645 *  0.0155 = 0.3745

Here we will reject the null hypothesis when p^ < 0.3745

(f) Here hypothesis is

H0: p = 0.4 vs. H1: p ≠ 0.4

at 1% significance level.

Here critical value = Z_critical = NORMSINV(0.995) = 2.576

so we would reject the null hypothesis if Z > 2.576 and Z < -2.576