Write (x+1)^5 in standard form?
2 answers:
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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:
<u>Step :1</u>
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
<u>Step 2</u>:-
<u>Null hypothesis </u>:- H₀ : x₁⁻ = x₂⁻
<u>Alternative hypothesis</u>:- H₁ : x₁⁻ ≠ x₂⁻
<u>level of significance</u> :- α= 0.05
The test statistic
where
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
s^2 = 1,205,391.42
Standard deviation = √1,205,391.42 = 1097.903
<u>Step 3</u>:-
The test statistic
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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