Elementary Statistics

Question

Given two proportions of males and females that recently completed their SATs, determine if these two proportions are significantly different from one another.

Male percentage above the 90th percentile is 84% out of 1190 males. Female Percentage above the 90th percentile is 86% out of 1345 females.

Provide the appropriate null and alternative hypotheses and reasons that demonstrate how your evidence either refutes or shows if significant differences exist.

Answer 1

n₁ = 1190 males

p₁ = 0.84 (males that scored above 90^th percentile on the SAT)

n₂ = 1345 females

p₂ = 0.86 (females that scored above 90^th percentile on the SAT)

Let's use the two-sided test.

H_o: p1 - p2 = 0

H_A: p1 - p2 ≠ 0

We will be using the 2-proportion z-test to determine if the two sample proportions are different. In addition, we will be combining the proportions (pooled estimate for proportions).

p_c = count of success in both samples combined ÷ count of students in both samples combined

p_c = (1190*0.84) + (1345*0.86) ÷ (1190 + 1345)

p_c = (1000 + 1157) ÷ (1190 + 1345)

p_c = 2157/2535 = 0.85

z = [(0.84 - 0.86) - 0] ÷ √[(0.85)(0.15)/1190] + [(0.85)(0.15)/1345]

z = -0.02 ÷ 0.01421 = -1.41

Use the z-score table to find the p-value.

p-value = 0.0793 > 0.05. Since the p-value is greater than the level of significance, α = 0.05, based on the assumption, we fail to reject the null hypothesis. There is no sufficient evidence that shows the two proportions of males and females that recently took the SATs are significantly different from one another.