Question

1. For each of the 2 majors, conduct a full hypothesis test at the 10% significance level:
   • The mean ‘Cost’ for a college is $160,000.
2. 2. For Business versus Engineering majors conduct a full, two-sample, full hypothesis test at the 5% significance level (assume the variances are not equal):The average ’30-Year ROI’ for Business majors is less than for Engineering Majors.

1. 3.  In a highlighted box, explain how each hypothesis test contributes to the central question of which major would give the better ROI. Why do we need hypothesis testing? Isn't it enough to look at the sample of 20 schools and look at those numbers? What exactly does hypothesis testing do that looking at the numbers in the sample and saying "one is higher than the other" cannot do? This answer is critical to your final project. It demonstrates that you understand why you just can't look at the mean, median, and mode of the spreadsheet and "call it a day."

   Business				Engineering
   Private	$222,700.00	$1,786,000.00	7.70%	Private	$221,700.00	$2,412,000.00	8.70%
   Private	$176,400.00	$1,758,000.00	8.40%	Private	$213,000.00	$2,064,000.00	8.30%
   Private	$212,200.00	$1,714,000.00	7.80%	Private	$230,100.00	$1,949,000.00	7.90%
   Public	$125,100.00	$1,535,000.00	9.10%	Private	$222,600.00	$1,947,000.00	8.00%
   Private	$212,700.00	$1,529,000.00	7.40%	Private	$225,800.00	$1,938,000.00	8.00%
   Public	$92,910.00	$1,501,000.00	10.10%	Public	$87,660.00	$1,937,000.00	11.20%
   Private	$214,900.00	$1,485,000.00	7.30%	Private	$224,900.00	$1,915,000.00	7.90%
   Private	$217,800.00	$1,483,000.00	7.20%	Private	$221,600.00	$1,878,000.00	7.90%
   Private	$225,600.00	$1,444,000.00	7.00%	Public	$125,100.00	$1,854,000.00	9.80%
   Private	$217,300.00	$1,442,000.00	7.10%	Private	$215,700.00	$1,794,000.00	7.90%
   Private	$226,500.00	$1,441,000.00	7.00%	Public	$92,530.00	$1,761,000.00	10.60%
   Private	$215,500.00	$1,438,000.00	7.20%	Private	$217,800.00	$1,752,000.00	7.70%
   Private	$223,500.00	$1,428,000.00	7.00%	Public	$89,700.00	$1,727,000.00	10.70%
   Private	$226,600.00	$1,414,000.00	7.00%	Private	$229,600.00	$1,716,000.00	7.50%
   Private	$189,300.00	$1,397,000.00	7.50%	Public	$101,500.00	$1,703,000.00	10.20%
   Public	$89,700.00	$1,382,000.00	9.90%	Public	$115,500.00	$1,694,000.00	9.70%
   Public	$87,030.00	$1,376,000.00	10.00%	Public	$104,500.00	$1,690,000.00	10.10%
   Private	$218,200.00	$1,343,000.00	6.90%	Public	$69,980.00	$1,685,000.00	11.50%
   Private	$229,900.00	$1,339,000.00	6.70%	Private	$219,400.00	$1,676,000.00	7.60%
   Private	$148,800.00	$1,321,000.00	8.10%	Public	$64,930.00	$1,668,000.00	11.70%

Answer 1

Sno	Business	Cost	Engineering	Cost
1	Private	222700	Private	221700
2	Private	176400	Private	213000
3	Private	212200	Private	230100
4	Public	125100	Private	222600
5	Private	212700	Private	225800
6	Public	92910	Public	87660
7	Private	214900	Private	224900
8	Private	217800	Private	221600
9	Private	225600	Public	125100
10	Private	217300	Private	215700
11	Private	226500	Public	92530
12	Private	215500	Private	217800
13	Private	223500	Public	89700
14	Private	226600	Private	229600
15	Private	189300	Public	101500
16	Public	89700	Public	115500
17	Public	87030	Public	104500
18	Private	218200	Public	69980
19	Private	229900	Private	219400
20	Private	148800	Public	64930
	Average	188632		164680
	SD	50503.43		66385.12

For Sample 1

The provided sample mean is X = 188632 and the sample standard deviation is 8 = 50503.43, and the sample size is n 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u 160000 Ha: 160000 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is a 0.10, and the critical value for a two- 1.729 tailed test is te The rejection region for this two-tailed test is R {t: \t| > 1.729} (3) Test Statistics The t-statistic is computed as follows: X — Но t = 188632 160000 2.535 50503.43/20 S

(4) Decision about the null hypothesis Since it is observed that t 2.535 > te = 1.729, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p 0.0202, and since p 0.0202 < 0.10, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 160000, at the 0.10 significance level.

For sample 2

The provided sample mean is X = 164680 and the sample standard deviation is s 66385.12, and the sample size is n = 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 160000 Ha: 160000 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a 0.10, and the critical value for a two- tailed test is te = 1.729. The rejection region for this two-tailed test is R {t : |t| > 1.729} (3) Test Statistics The t-statistic is computed as follows: X t 164680 160000 = 0.315 66385.12/20 S

(4) Decision about the null hypothesis Since it is observed that t 0.315 < te = 1.729, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.756, and since p = 0.756 > 0.10, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 160000, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 139012.469 < u < 190347.531.

Question 2

The provided sample means are shown below: XI 188632 X2 164680 Also, the provided sample standard deviations are: 50503.43 S 66385.12 S2 20 and n2 20 and the sample sizes are n (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 2 Ha: 2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a 0.05, and the degrees of freedom are df 38. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is te 2.024, for a 0.05 and df 38 The rejection region for this two-tailed test is R {t : \t| > 2.024}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: (21-1)(n2-)s + n1+n2-2 188632 164680 1.284 (20-1)50503.432+(20-1)66385.122 20+20-2 1 20 20 (4) Decision about the null hypothesis Since it is observed that t= 1.284 < te = 2.024, it is then concluded that the null hypothesis is not rejected

Using the P-value approach: The p-value is p 0.2069, and since p 0.2069 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 2, at the 0.05 significance level. Confidence Interval The 95% confidence interval is -13806.019 < - <61710.019