Which equation could have been used to create this graph?
2 answers:
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Answer:
a) Private Colleges
Sample mean = 42.5 thousand dollars
Standard deviation = S1 = 6.62 thousand dollars.
Public colleges
Sample mean = 22.3 thousand dollars
Standard deviation = 4.34 thousand dollars
b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars
The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to public colleges
c) 95% confidence interval for a sampling distribution of the difference of the cost of private and public colleges is given as
(15.0, 25.4) thousand dollars.
Step-by-step explanation:
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) Calculate sample mean and standard deviation for both data set.
Mean = (Σx)/N
where N = Sample size
Σx = sum of all variables
Private colleges
Σx = (52.8+43.2+45.0+33.3+44.0+30.6+45.8+37.8+50.5+42.0) = 425
N = 10
Mean = 425/10 = 42.5 thousand dollars
Standard deviation = S1 = √[Σ(x - xbar)²/N]
Σ(x - xbar)² = (52.8-42.5)² + (43.2-42.5)²
+ (45.0-42.5)² + (33.3-42.5)²
+ (44.0-42.5)² + (30.6-42.5)² + (45.8-42.5)² + (37.8-42.5)² + (50.5-42.5)² + (42.0-42.5)² = 438.56
N = 10
Standard deviation = √[Σ(x - xbar)²/N]
Standard deviation = √(438.56/10) = 6.62 thousand dollars
Public colleges
Mean = (Σx)/N
Σx =
(20.3+22.0+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) = 267.6
N = 12
Mean = (267.6/12) = 22.3 thousand dollars
Standard deviation = √[Σ(x - xbar)²/N]
[Σ(x - xbar)²
(20.3-22.3)² + (22.0-22.3)² + (28.2-22.3)² + (15.6-22.3)² + (24.1-22.3)² + (28.5-22.3)² + (22.8-22.3)² + (25.8-22.3)² + (18.5- 22.3)² + (25.6-22.3)² +(14.4-22.3)+(21.8-22.3) = 225.96
N = 12
standard deviation = s2 = √(225.96/12) = 4.34 thousand dollars
b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars
The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to public colleges.
c. Develop a 95% confidence interval of the difference between the annual cost of attending private and pubic colleges.
95% confidence interval, private colleges have a population mean annual cost $ to $ more expensive than public colleges.
To combine the distribution in this manner,
Sample mean of difference = 20.2 thousand dollars
Combined standard deviation of the sampling distribution = √[(S1²/n1) + (S2²/n2)]
= √[(6.62²/10) + (4.34²/12)] = 2.44 thousand dollars
Confidence interval = (Sample mean) ± (Margin of error)
Sample mean = 20.2
Margin of error = (critical value) × (standard deviation of the sampling distribution)
standard deviation of the sampling distribution = 2.44
To obtain the critical value, we need the t-score at a significance level of 5%; α/2 = 0.025
we obtain the degree of freedom too
The degree of freedom, df, is calculated in the attached image.
df = 15
t (0.025, 15) = 2.13145 from the tables
Margin of error = 2.13145 × 2.44 = 5.20
Confidence interval = (Sample mean) ± (Margin of error)
= (20.2 ± 5.2) = (15.0, 25.4)
Hope this Helps!!!
