Question

City health officials will conduct a two-sample t-test for a difference in means to investigate whether

Answer 1

Answer:

C) t=120−100/√(20^2/22)+(30^2/15)

Step-by-step explanation:

The numerator is the difference in the sample means, or 120−100. The denominator is the standard error of the difference in sample means, or √(20^2/22)+(30^2/15)

Answer 2

Answer:

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means      

Step-by-step explanation:

Step(i):-

Given that the first sample size 'n₁' = 22

Given that the mean of the first sample x₁⁻ = 120

Given that the standard deviation of the sample (s₁) = 20

Given that the mean of the second sample x₂⁻ = 100

Given that the standard deviation of the second sample (S₂) = 30

Step(ii):-

T-test statistic

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

     where

             $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2} S_{2} ^{2} }{n_{1} +n_{2}-2 }$

            $S^{2} = \frac{22( 20)^{2}+ 15(30) ^{2} }{22 +15-2 }$

           S² =  637.1428

           S = √637.1428 = 25.24168  

The standard deviation of the sample 'S' = 25.24168   

Step(iii):-

Null Hypothesis:H₀:

There is no difference between the means

Alternative Hypothesis:H₁:

There is a difference between the means

         T-test statistic

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

              $t = \frac{ 120 -100 }{\sqrt{637.14 (\frac{1}{22 }+\frac{1}{15 } ) } }$

             t =   2.366

Degrees of freedom

           γ = n₁+n₂ -2 = 22+15-2 = 35

     t₀.₀₅ = 2.0301

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means