1. A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners. The mean for the runners was 75.9 mm Hg with an SS of 1,500, and the mean for the nonrunners was 80.3 mm Hg with an SS of 8. Is running a good way to lower a person's blood pressure? a. Test the null hypothesis at the .05 level

Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

running a good way to lower a person's blood pressure

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

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

Step 3:-

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.

running a good way to lower a person's blood pressure

2.83941151 would be the answer you're looking for! : )

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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