Answer: 0.07 to 1.9
Step-by-step explanation:
Let [tex]\mu_1[/tex] and [tex]\mu_2[/tex] be the population mean of hourly wages of employees of two branches of a department store (Downtown Store and North Mall Store)
As per given , we have
[tex]n_1=25\\\\ n_2=20[/tex]
[tex]\overline{x}_1=\$9\\\\\overline{x}_2=\$8[/tex]
[tex]s_1=\$2\\\\s_1=\$1[/tex]
Degree of freedom : [tex]\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s _2^2}{n_2})^2}{\dfrac{(\dfrac{s_1^2}{n_1})^2}{n_1-1}+\dfrac{(\dfrac{s_2^2}{n_2})^2}{n_2-1}}[/tex]
[tex]=\dfrac{(\dfrac{2^2}{25}+\dfrac{1^2}{20})^2}{\dfrac{(\dfrac{2^2}{25})^2}{24}+\dfrac{(\dfrac{1^2}{20})^2}{19}} =36[/tex]
Critical value for 95% confidence interval :
[tex]t_{\alpha/2, df}=t_{0.025,\ 36}=2.028[/tex]
Then, 95% interval estimate for the difference between the two population means will be
[tex]\overline{x}_1-\overline{x}_2\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s _2^2}{n_2}}[/tex]
[tex]9-8\pm (2.028)\sqrt{\dfrac{2^2}{25}+\dfrac{1^2}{20}}[/tex]
[tex]1\pm 0.93=(1-0.93,\ 1+0.93 ) = (0.07,\ 1.93)[/tex]
Hence, A 95% interval estimate for the difference between the two population means is [tex](0.07, 1.93)[/tex] .
i.e. 0.07 to 1.9
