Question

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

Answer 1

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