An article presents a study of health outcomes in women with symptoms of heart disease. In a sample of 115 women whose test results suggested the presence of coronary artery disease, the mean peak systolic blood pressure was 169.9 mmHg, with a standard deviation of 24.8 mmHg. In a sample of 235 women whose test results suggested an absence of coronary artery disease, the mean peak systolic blood pressure was 163.3 mmHg, with a standard deviation of 25.8 mmHg. Can you conclude that the mean peak systolic blood pressure differs between these two groups of women

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jtellezd jtellezd · Answer 1 · 2021-04-22T05:00:43+02:00

Answer:

z(s) is in the rejection region we reject H₀ μ₁ = μ₂ and support the claim that at CI 95 % the means of the two groups differs

Step-by-step explanation:

Sample 1:

Sise sample n₁ = 115

μ₁ = 169,9 mmHg

σ₁ = 24,8 mmHg

Sample 2:

Sise sample n₂ = 235

μ₂ = 163,3 mmHg

σ₂ = 25,8 mmHg

We can develop a test hypothesis for differences in means to investigate if the mean peak systolic blood pressure differs between these two groups

We will choose CI = 95 % then significance level α = 5 %

α = 0,05 α/2 = 0,025

z(c) for 0,025 is from z-table z(c) = 1,96

Test Hypothesis:

Null Hypothesis H₀ μ₁ = μ₂

Alternative Hypothesis Hₐ μ₁ ≠ μ₂

The alternative hypothesis tells us that the test is a two-tail test.

z(s) = ( μ₁ - μ₂ ) / √ σ₁²/n₁ + σ₂²/n₂

z(s) = ( 169,9 -163,3 ) / √ (24,8)² /115 + ( 25,8)²/235

z(s) = 6,6 / √5,35 + 2,83

z(s) = 6,6 / 2,86

z(s) = 2,30

Comparing |z(c)| and |z(s)|

z(s) > z(c)

z(s) is in the rejection region we reject H₀ μ₁ = μ₂ and support the claim that at CI 95 % the means of the two groups differs