Answer :
Answer:
option C
Step-by-step explanation:
given,
[tex]\bar{x_1} = 158.706, \sigma_1 = 25.36 , n_1= 17[/tex]
[tex]\bar{x_1} = 122.471, \sigma_1 = 25.183 , n_2= 17[/tex]
α = 1 - 0.95 = 0.05
degree of freedom (df) = 17 -1 = 16
critical value[tex]= t_{\alpha/2},df = t_{0.025},16 =2.120[/tex] (from t-table)
margin of error = [tex]t_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}[/tex]
=2.120\times \sqrt{\dfrac{25.36^2}{17}+\dfrac{25.183^2}{17}}[/tex]
= 2.120 x 8.6467
= 18.33
Margin of error = 18.33
Point estimation of difference = [tex]\bar{x_1} - \bar{x_2}[/tex]
= 36.235
lower limit = 36.235 - 18.33 = 17.91
upper limit = 36.235 + 18.33 = 54.57
hence, the nearest option near to answer is option C