Answer: [tex](0.85,\ 3.15)[/tex]
Step-by-step explanation:
Given : [tex]n_x=12\ ;\ \mu_x=4.8\ ;\ \sigma_x=1.9[/tex]
[tex]n_y=24\ ;\ \mu_y=2.8\ ;\ \sigma_y=1.0[/tex]
Significance level : [tex]\alpha: 1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Now, the confidence interval for difference of two population mean :-
[tex]\mu_x-\mu_y\pm z_{\alpha/2}(\sqrt{\dfrac{\sigma_x^2}{n_x}+\dfrac{\sigma_y^2}{n_y}}\\\\=4.8-2.8\pm(1.96)(\sqrt{\dfrac{(1.9)^2}{12}+\dfrac{(1.0)^2}{24}}\\\\\approx2\pm1.15\\\\=(0.85,\ 3.15)[/tex]
Hence, the 95% confidence interval for the difference [tex]\mu_x-\mu_y=(0.85,\ 3.15)[/tex]
