Answer:
The correct 90% confidence interval for the difference between women and men in the average number of hours of personal time taken per year is
contained by the interval (0.54 , 5.18)
Step-by-step explanation:
Given that:
for women
sample size = n₁ = 16
sample mean [tex](\bar X_1)[/tex] = 24.75
standard deviation s₁ = 2.84
for men
sample size for men = n₂ = 7
sample mean [tex](\bar X_2)[/tex] = 21.89
standard deviation s₂ = 3.29
degree of freedom = [tex](n_1 + n_2 )-2[/tex]
= 16 + 7 -2
= 21
The critical value for [tex]\alpha = 0.1 \ and \ df \ of \ 21 \ is :[/tex]
[tex]t_{0.005;21}= 1.721[/tex]
The pooled standard deviation can be calculated as follows since the population variances are assumed to be equivalent.
[tex]s_p = \sqrt{\dfrac{(n_1 -1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}[/tex]
[tex]s_p = \sqrt{\dfrac{(16 -1)2.84^2+(7-1)3.29^2}{16+7-2}}[/tex]
[tex]s_p = \sqrt{\dfrac{(15)8.0656+(6)10.8241}{21}}[/tex]
[tex]s_p = 2.976[/tex]
The standard error is estimated as follow:
[tex]se = s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}[/tex]
[tex]se = 2.976\sqrt{\dfrac{1}{16} + \dfrac{1}{7}}[/tex]
se = 1.348
Finally, the confidence interval [tex]C.I = ( x_1 -x_2 -t_c \times se \ \ ; \ \ x_1-x_2 +t_c \times se)[/tex]
[tex]=(24.75 - 21.89 -1.721 \times 1.348 \ \ , \ \ 2475 - 21.89 + 1.721 \times 1.348) \\ \\ \mathbf{ = (0.54 \ \ , \ \ 5.18)}[/tex]
