Respuesta :
Answer:
(a) Sample mean for private and public colleges are 42.5 and 22.3 respectively.
Sample standard deviation for private and public colleges are 4.98 and 4.53 respectively.
(b) The point estimate of the difference between the two population means is 20.3.
(c) 95% confidence interval of the difference between the annual cost of attending private and public colleges is [$15.05 , $25.35].
Step-by-step explanation:
We are given the following random samples which shows the annual cost of attending private and public colleges;
Private colleges.
52.8, 43.2, 45.0, 33.3, 44.0, 30.6, 45.8, 37.8, 50.5, 42.0
Public colleges.
20.3, 22.0, 28.2, 15.6, 24.1, 28.5, 22.8, 25.8, 18.5, 25.6, 14.4, 21.8
(a) Sample mean for private colleges, [tex]\bar X_1[/tex] = [tex]\frac{\sum X_i}{n}[/tex]
[tex]\bar X_1 = \frac{52.8+ 43.2+ 45.0+ 33.3+ 44.0 +30.6+ 45.8+ 37.8+ 50.5+ 42.0}{10}[/tex]
[tex]\bar X_1[/tex] = 42.5
Also, Sample standard deviation for private colleges, [tex]s_1[/tex] = [tex]\sqrt{\frac{\sum (X_i-\bar X_1)^{2} }{n-1} }[/tex]
So, [tex]s_1[/tex] = 6.98
Similarly, Sample mean for public colleges, [tex]\bar X_2[/tex] = [tex]\frac{\sum X_i}{n}[/tex]
[tex]\bar X_2 = \frac{20.3+ 22.0+ 28.2+ 15.6+ 24.1+ 28.5+ 22.8+ 25.8+ 18.5+ 25.6 +14.4+ 21.8}{12}[/tex]
[tex]\bar X_2[/tex] = 22.3
Also, Sample standard deviation for public colleges, [tex]s_2[/tex] = [tex]\sqrt{\frac{\sum (X_i-\bar X_2)^{2} }{n-1} }[/tex]
So, [tex]s_2[/tex] = 4.53
(b) The point estimate of the difference between the two population means = [tex]\bar X_1[/tex] - [tex]\bar X_2[/tex] = 42.5 - 22.3 = 20.2
The interpretation of this value is that the difference in population mean annual cost of attending private and public colleges is $20,200.
(c) The Pivotal quantity for 95% confidence interval of the difference between the population means is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1-_n__2-2[/tex]
where, [tex]\mu_1[/tex] = population mean cost of attending private colleges
[tex]\mu_2[/tex] = population mean cost of attending public colleges
[tex]n_1[/tex] = sample of private colleges data = 10
[tex]n_1[/tex] = sample of public colleges data = 12
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(10-1)\times 6.98^{2} +(12-1)\times 4.53^{2} }{10+12-2} }[/tex] = 5.763
Here for constructing 95% confidence interval we have used Two-sample t test statistics.
So, 95% confidence interval for the difference between population means ([tex]\mu_1-\mu_2[/tex]) is ;
P(-2.086 < [tex]t_2_0[/tex] < 2.086) = 0.95 {As the critical value of t at 20 degree
of freedom are -2.086 & 2.086 with P = 2.5%}
P(-2.086 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < 2.086) = 0.95
P( [tex]-2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ) = 0.95
P( [tex](\bar X_1-\bar X_2)-2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ) = 0.95
95% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =
[ [tex](\bar X_1-\bar X_2)-2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+2.086 \times }{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ]
=[[tex](42.5-22.3)-2.086 \times }{5.763\sqrt{\frac{1}{10} +\frac{1}{12} } }[/tex] ,[tex](42.5-22.3)+2.086 \times }{5.763\sqrt{\frac{1}{10} +\frac{1}{12} } }[/tex]]
= [15.05 , 25.35]
Therefore, 95% confidence interval of the difference between the annual cost of attending private and public colleges is [$15.05 , $25.35].
