Question

1. You now no longer believe that the population standard deviation in the tuition amounts charged by this population of universities is a known quantity. You therefore will use the sample standard deviation of the tuition amounts as an estimate of this unknown population standard deviation as you attempt to estimate the mean tuition of this population of universities. You collect tuition information from a random sample of universities. This data is shown in appendix one below. Find both 95% and 90% confidence intervals for the mean yearly tuition of this population of universities. Explain the meaning of these intervals. At each confidence level, comment upon the change in the results of this problem from the results of the previous problem.

           Appendix One (Yearly Tuition)

           $15,678        $21,409        $32,311           $20,945        $22,375        $10,456           $25,674        $35,151        $18,904           $30,500

           $23,571        $28,612        $21,786           $17,776        $18,905

           $23,678        $25,892        $21,090           $16,736        $19,999

           $21,689        $36,792        $23,893           $15,430        $22,229

           $26,903        $21,378        $26,500           $30,000        $22,287

           $38,017        $23,765        $21,001           $28,739        $25,000

           $23,001`       $24,590        $21,017           $17,450        $19,540

This assignment is designed to illustrate how a software package such as Microsoft Excel supplemented by an add-in such as PHStat can enable one to make decisions about the behavior of a single population parameter. You must provide printouts of all results of the hypothesis tests that you accomplish as you fulfill the requirements of this assignment. You should not only find the required information to make the appropriate decisions, but you also should provide explanations, in the context of the problems, of the meanings of your decisions.

Answer 1

Cofidence Interval =     , at ( n-1 ) degree of freedom.

S.No.	X	X-Mean(X)	[X-Mean(X)]^2
1	15678	-7838.725	61445609.63
2	21409	-2107.725	4442504.676
3	32311	8794.275	77339272.78
4	20945	-2571.725	6613769.476
5	22375	-1141.725	1303535.976
6	10456	-13060.725	170582537.5
7	25674	2157.275	4653835.426
8	35151	11634.275	135356354.8
9	18904	-4612.725	21277231.93
10	30500	6983.275	48766129.73
11	23571	54.275	2945.775625
12	28612	5095.275	25961827.33
13	21786	-1730.725	2995409.026
14	17776	-5740.725	32955923.53
15	18905	-4611.725	21268007.48
16	23678	161.275	26009.62563
17	25892	2375.275	5641931.326
18	21090	-2426.725	5888994.226
19	16736	-6780.725	45978231.53
20	19999	-3517.725	12374389.18
21	21689	-1827.725	3340578.676
22	36792	13275.275	176232926.3
23	23893	376.275	141582.8756
24	15430	-8086.725	65395121.23
25	22229	-1287.725	1658235.676
26	26903	3386.275	11466858.38
27	21378	-2138.725	4574144.626
28	26500	2983.275	8899929.726
29	30000	6483.275	42032854.73
30	22287	-1229.725	1512223.576
31	38017	14500.275	210257975.1
32	23765	248.275	61640.47563
33	21001	-2515.725	6328872.276
34	28739	5222.275	27272156.18
35	25000	1483.275	2200104.726
36	23001	-515.725	265972.2756
37	24590	1073.275	1151919.226
38	21017	-2499.725	6248625.076
39	17450	-6066.725	36805152.23
40	19540	-3976.725	15814341.73
		Total	1306535666

Sample Mean=	23516.725
Sample Variance=	2717049389626930.00
Sample Standard Deviation=	52125323.88

Sample Mean =

Sample Variance =

Sample Standard Deviation = Sqrt( Sample Variance )

(i) Confidence Interval at 95% i.e. ,

C.I = 23516.725 ( 2.021 )*(521256323.88) / Sqrt(40)

     =   23516.725 (2.021)*(521256323.88)/ (6.3245)

     = 23516.725 166567954.867

     = ( -16633238.142 , 166591471.592 )

(ii) Confidence Interval at 90% i.e. ,

C.I = 23516.725 ( 1.684 )*(521256323.88) / Sqrt(40)

     = 23516.725 ( 1.684 ) * (521256323.88) / (6.3245)

     = 23516.75 138792892.625

     = ( - 138769375.875 , 138816409.375 )

Here, decision based on testing of hypothesis is not possible because the data for Population Mean is not given.