Respuesta :
Answer:
Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.
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Step-by-step explanation:
Step :1
A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners
The first sample size is n₁ =15
The second sample size is n₂ =15
Given data the mean for the runners was 75.9 mm Hg with an SS of 1,500
The mean of the first sample x₁⁻ = 75.9 mm
The mean of the second sample x₂⁻ = 80.3mm
The standard deviation of the first sample (S₁) = 1,500
The standard deviation of the second sample (S₂) = 8
Step 2:-
Null hypothesis :- H₀ : x₁⁻ = x₂⁻
Alternative hypothesis:- H₁ : x₁⁻ ≠ x₂⁻
level of significance :- α= 0.05
The test statistic [tex]t = \frac{x_{1}^- -x_{2}^- }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }[/tex]
where [tex]S^{2} =\frac{n_{1}S_{1} ^2+n_{2}S_{2} ^2 }{n_{1}+n_{2}-2}[/tex]
n₁ =15 ,n₂ =15 x₁⁻ = 75.9 mm ,x₂⁻ = 80.3mm and (S₁) = 1,500 and (S₂) = 8
substitute all values in above equation, we get
[tex]S^{2} =\frac{15X(1500) ^2+15X(8) ^2 }{15+15-2}[/tex]
s^2 = 1,205,391.42
Standard deviation = √1,205,391.42 = 1097.903
Step 3:-
The test statistic
[tex]t = \frac{x_{1}^- -x_{2}^- }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }[/tex]
x₁⁻ = 75.9 mm ,x₂⁻ = 80.3mm, n₁ =15 ,n₂ =15 and S = 1097.903
The test statistic value t = -0.01097
modulus t = 0.0109
Calculated value t =0.0109
The degrees of freedom γ=n₁+n₂ -2 = 15+15 -2 =28
From t- distribution table
From tabulated value t = 1.701 at 28 degrees of freedom at 0.05 level of significance.
Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.
Therefore we accepted null hypothesis.
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