Question

In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows:
Brand X n2=20 xbar 2=6.80 s2=1.15
Brand Y n1=20 xbar1=7.30 s1=1.10
Assume that the population variances are equal. Then, the standard error of the sampling distribution of the sample mean difference xbar1−xbar2 is equal to: Question 2 options: (a) 1.1275 (b) 0.1266 (c) 1.2663 (d) 0.3558.

Answer 1

Answer:  (d) 0.3558.

Step-by-step explanation:

We know that the standard error of sample mean difference is given by:-

$S.E.=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}$

Given : $n_1= 20\ ,\ n_2=20$

$s_1=1.10\ ,\ \ s_2=1.15$

Then , the standard error of the sampling distribution of the sample mean difference $\overline{x_1}-\overline{x_2}$ is equal to :-

$S.E.=\sqrt{\dfrac{1.10^2}{20}+\dfrac{1.15^2}{20}}\\\\\Rightarrow\ S.E.=0.355844066973\approx0.3558$

Hence, the standard error of the sampling distribution of the sample mean difference $\overline{x_1}-\overline{x_2}$ is equal to 0.3558.