Question

A researcher wishes to determine whether people with high blood pressure can lower their blood pressure by performing yoga exercises. A treatment group and a control group are selected. The sample statistics are given below. Construct a 90% confidence interval for the difference between the two population means, Would you recommend using yoga exercises? Treatment Group Control Group n1 = 100 n2 = 100 1 = 178 2 = 193 s1 = 35 s2 = 37

Answer 1

Answer:

90% confidence interval for the difference between the two population means

( -23.4166 , -6.5834)

Step-by-step explanation:

Step(i):-

Given first sample size n₁ = 100

Given mean of the first sample x₁⁻ = 178

Standard deviation of the sample S₁ = 35

Given second sample size n₂= 100

Given mean of the second sample x₂⁻ = 193

Standard deviation of the sample S₂ = 37

Step(ii):-

Standard error of two population means

        $se(x^{-} _{1} -x^{-} _{2} ) = \sqrt{\frac{s^{2} _{1} }{n_{1} }+\frac{s^{2} _{2} }{n_{2} } }$

       $se(x^{-} _{1} -x^{-} _{2} ) = \sqrt{\frac{(35)^{2} }{100 }+\frac{(37)^{2} }{100 } }$

        $se(x^{-} _{1} -x^{-} _{2} ) = 5.093$

Degrees of freedom

ν  = n₁ +n₂ -2 = 100 +100 -2 = 198

t₀.₁₀ = 1.6526

Step(iii):-

90% confidence interval for the difference between the two population means

$(x^{-} _{1} - x^{-} _{2} - t_{\frac{\alpha }{2} } Se (x^{-} _{1} - x^{-} _{2}) , x^{-} _{1} - x^{-} _{2} + t_{\frac{\alpha }{2} } Se (x^{-} _{1} - x^{-} _{2})$

(178-193 - 1.6526 (5.093) , 178-193 + 1.6526 (5.093)

(-15-8.4166 , -15 + 8.4166)

( -23.4166 , -6.5834)