Question

The UWGB student body is 67% female. If a room of students consists of 42 females and 18 males, is this room of students a good fit with the UWGB student body. Test using an alpha of 0.01.

a) State the null and alternate hypotheses in symbols if appropriate and words.

B) The test you will use and why.

C) sketch the distribution showing the rejection region.

D) calculate the test statistic or p-value.

E) determine whether you reject or fail to reject the null hypothesis and justify your decision.

f) interpret your decision in the context of the problem.

Answer 1

Random sample (n) = 70

Number of female in a room (x) = 42

= x/n = 42/70= 0.60

1. The hypotheses are

H₀: p = 0.67   The UWGB student body is 67% female.

H_a: p ≠ 0.67   The UWGB student body is not 67% female.

2. We use z test for one population proportion test because we have given the values of population proportion.
3. Two sided z critical values at α = 0.01 is Z_α/2 = Z_0.005 = -2.58, 2.58

The rejection region in blue colour.

4. Test statistics: Z = ( – P₀) / √ (P₀ (1 - P₀) / n)

                        Z = (0.60 – 0.67) / √(0.67)*(1-0.67)/70

                           Z = -1.246

            P value =2* P( z > 1.246) = 2*(1-NORM.S.DIST(1.246,TRUE))= 0.213

5. Decision rule: Reject H₀ if |Z| > Z₀. Here |Z| = 1.246 < Z₀ = 2.58 and P-value = 0.213 > 0.01 Since test statistics not lies in rejection region so we failed to reject null hypothesis.
6. Conclusion: Failed to reject H₀. There is not enough evidence to reject the claim that the UWGB student body is 67% female.