Question

Test the Business Major's hypothesis at the 5% significance level. The mean ‘Cost’ for a college is $160,000. Be sure to interpret your results.

• List your givens - n, standard deviation, alpha...
• Provide the formula you will use. Why are you using that formula? How does it fit the situation?
• Show how you have plugged in the numbers.
• What is your computed test statistic?
• Show the computation of the p value.
• Will you reject or fail to reject the null hypothesis. Why?

School Type	Cost	30 Year ROI	Annual
Private	$222,700.00	$1,786,000.00	7.70%
Private	$176,400.00	$1,758,000.00	8.40%
Private	$212,200.00	$1,714,000.00	7.80%
Public	$125,100.00	$1,535,000.00	9.10%
Private	$212,700.00	$1,529,000.00	7.40%
Public	$92,910.00	$1,501,000.00	10.10%
Private	$214,900.00	$1,485,000.00	7.30%
Private	$217,800.00	$1,483,000.00	7.20%
Private	$225,600.00	$1,444,000.00	7.00%
Private	$217,300.00	$1,442,000.00	7.10%
Private	$226,500.00	$1,441,000.00	7.00%
Private	$215,500.00	$1,438,000.00	7.20%
Private	$223,500.00	$1,428,000.00	7.00%
Private	$226,600.00	$1,414,000.00	7.00%
Private	$189,300.00	$1,397,000.00	7.50%
Public	$89,700.00	$1,382,000.00	9.90%
Public	$87,030.00	$1,376,000.00	10.00%
Private	$218,200.00	$1,343,000.00	6.90%
Private	$229,900.00	$1,339,000.00	6.70%
Private	$148,800.00	$1,321,000.00	8.10%

Answer 1

data

cost
222700
176400
212200
125100
212700
92910
214900
217800
225600
217300
226500
215500
223500
226600
189300
89700
87030
218200
229900
148800

One-Sample T: cost

Test of μ = 160000 vs ≠ 160000

Variable   N    Mean StDev SE Mean       95% CI          T      P
cost      20 188632 50503    11293 (164996, 212268) 2.54 0.020

List your givens - n, standard deviation, alpha...

n = 20 , sd = 50503 , alpha = 0.05 ,xbar = 188632, mu = 160000

Provide the formula you will use. Why are you using that formula? How does it fit the situation?

TS = (xbar - mu)/(sd/sqrt(n))

What is your computed test statistic?

TS = (xbar - mu)/(sd/sqrt(n))

= 2.54

Show the computation of the p value.

p-value = 2 P(T > 2.54) = 0.020

Will you reject or fail to reject the null hypothesis. Why?

p-value < 0.05

hence we reject the null hypothesis