In testing whether the means of two normal populations are equal,summary statistics computed for two independent samples are as follows: $$\begin{array} { | c | c | }
\hline \text { Brand } \boldsymbol { X } & \text { Brand } \boldsymbol { Y } \
\hline n _ { 1 } = 25 & n _ { 2 } = 25 \
\bar { x } _ { 1 } = 7.30 & \bar { X } _ { 2 } = 6.80 \
s _ { 1 } = 1.05 & s _ { 2 } = 1.20 \
\hline
\end{array}$$ Assume that the population variances are equal.Then,the standard error of the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ is equal to:
A) 0.1017.
B) 1.2713.
C) 0.3189.
D) 1.1275.

Question

In testing whether the means of two normal populations are equal,summary statistics computed for two independent samples are as follows: $$\begin{array} { | c | c | } 
\hline \text { Brand } \boldsymbol { X } & \text { Brand } \boldsymbol { Y } \
\hline n _ { 1 } = 25 & n _ { 2 } = 25 \
\bar { x } _ { 1 } = 7.30 & \bar { X } _ { 2 } = 6.80 \
s _ { 1 } = 1.05 & s _ { 2 } = 1.20 \
\hline
\end{array}$$ Assume that the population variances are equal.Then,the standard error of the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ is equal to:
A) 0.1017.
B) 1.2713.
C) 0.3189.
D) 1.1275.

Quizplus · Accepted Answer

The standard error of the difference between two means is calculated using the formula: $ SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} $, where $ s_p^2 $ is the pooled variance. Here, $ s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{24 \times 1.05^2 + 24 \times 1.20^2}{48} = 1.1275 $. Thus, $ SE = \sqrt{\frac{1.1275}{25} + \frac{1.1275}{25}} = 0.3189 $.

In Testing Whether the Means of Two Normal Populations Are