MATH SOLVE

4 months ago

Q:
# 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=1.15Brand 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.

Accepted Solution

A:

Answer: (d) 0.3558.Step-by-step explanation:We know that the standard error of sample mean difference is given by:-[tex]S.E.=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}[/tex]Given : [tex]n_1= 20\ ,\ n_2=20[/tex][tex]s_1=1.10\ ,\ \ s_2=1.15[/tex]Then , the standard error of the sampling distribution of the sample mean difference [tex]\overline{x_1}-\overline{x_2}[/tex] is equal to :-[tex]S.E.=\sqrt{\dfrac{1.10^2}{20}+\dfrac{1.15^2}{20}}\\\\\Rightarrow\ S.E.=0.355844066973\approx0.3558[/tex]Hence, the standard error of the sampling distribution of the sample mean difference [tex]\overline{x_1}-\overline{x_2}[/tex] is equal to 0.3558.