In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows: n₁ = 25, $\bar { x } _ { 1 }$ = 7.30, s₁ = 1.05.
N₂ = 30, $\bar { x } _ { 2 }$ = 6.80, s₂ = 1.20.
Assume that the population variances are equal. Then the standard error of the sampling distribution of the sample mean difference $\bar { x } _ { 1 } - \bar { x } _ { 2 }$ is equal to:

Question

In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows: n₁ = 25,

\bar { x } _ { 1 }

= 7.30, s₁ = 1.05.
N₂ = 30,

\bar { x } _ { 2 }

= 6.80, s₂ = 1.20.
Assume that the population variances are equal. Then the standard error of the sampling distribution of the sample mean difference

\bar { x } _ { 1 } - \bar { x } _ { 2 }

is equal to:

Quizplus · Accepted Answer

The standard error of the sampling distribution of the sample mean difference $\bar { x } _ { 1 } - \bar { x } _ { 2 }$ is equal to:$$ \sqrt { \frac { \sigma ^ { 2 } } { n _ { 1 } } + \frac { \sigma ^ { 2 } } { n _ { 2 } } } = \sqrt { \frac { s ^ { 2 } } { n _ { 1 } } + \frac { s ^ { 2 } } { n _ { 2 } } } $$where s^2 is the pooled variance, which is estimated by:$$ s ^ { 2 } = \frac { ( n _ { 1 } - 1 ) s _ { 1 } ^ { 2 } + ( n _ { 2 } - 1 ) s _ { 2 } ^ { 2 } } { n _ { 1 } + n _ { 2 } - 2 } $$Plugging in the given values, we get:$$ s ^ { 2 } = \frac { ( 25 - 1 ) ( 1.05 ) ^ { 2 } + ( 30 - 1 ) ( 1.20 ) ^ { 2 } } { 25 + 30 - 2 } = 1.125 $$Therefore, the standard error of the sampling distribution of the sample mean difference is:$$ \sqrt { \frac { 1.125 } { 25 } + \frac { 1.125 } { 30 } } = 0.3072 $$

In Testing Whether the Means of Two Normal Populations Are xˉ1\bar { x } _ { 1 }xˉ1​

In Testing Whether the Means of Two Normal Populations Are $\bar { x } _ { 1 }$