the cost of education in western united states compared to southeastern united states

7/30/2019 The Cost of Education in Western United States Compared to Southeastern United States

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Economic Statistics

Probability Exploration

Project 3: November 8, 2012

The Cost of Education in Western United States

Compared to Southeastern United States

Ruth Clements, Kellen Sanger, Yakov Kagan


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Introduction

The United States is experiencing an ever increasing cost-of-living. While inflation and

CPI remain stagnant, it is clear that it is becoming more and more difficult to have a comfortable

lifestyle in the US. Coming from disparate parts of the country, we have noticed that the South is

relatively inexpensive comparative to other parts of the nation. As college students at a

prestigious, expensive, private university with a strong southern influence, we would like to

compare the cost of educationwhat we believe to be a good barometer for cost-of-livingin

two different regions of the country. According to Colleges freeze, reduce tuition as public

balks at further price hikes

1

, some schools in the southern part of the US, such as Sewanee have

cut tuition in order to retain students. In contrast, the University of California has raised tuition

for the past two years according to UC plan sees tuition rising up to 16% annually over four

years.2

We would like to compare the cost of college tuition in the southeastern part of the

United States as compared to the western part. In order to do so, we took a random sample of

schools in the South, and a random sample of schools in the West. We selected the schools in

each regions respective football Division I conferences. We would like to test our hypothesis

that the South has a lower cost of education than the West.

1http://usnews.nbcnews.com/_news/2012/08/01/13070188-colleges-freeze-reduce-tuition-as-

public-balks-at-further-price-hikes?lite2http://articles.latimes.com/2011/sep/15/local/la-me-0915-uc-plan-20110915


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Analysis

In mathematical notation, we defined our hypothesis that West coast schools cost more

than Southeastern schools as H0: 1 - 2 = 0 for the null and H1: 12 > 0 for the alternative.

H0: 1 - 2 = 0H1: 12 > 0

The population mean cost of all West coast schools is represented by 1 and the population

mean cost of all Southeastern schools is represented by 2. The hypothesis test we have

constructed tests the difference between these two means. By not rejecting the null, we will

conclude that the mean cost of all West coast schools is equal to the mean cost of all southeastern

schools. If the null is rejected in favor of the alternative, the test will indicate that the cost of the

West coast schools is greater than the cost of the Southeastern schools as denoted by the

difference being > 0.

In order to proceed with our hypothesis test of the difference between the population

means, we had to determine which t-test of u1-u2 to use. We conducted an F test using the

sample variances to do so. The F test tests the ratio of the two population variances to determine

if they are equal or not. Our null and alternative hypotheses were:

H0: 12/ 22 = 1

H1: 12/ 2

2 1

To calculate the F test statistic we used the sample variances that we obtained from our data. The

F test statistic can be calculated by: F= s12/s2

2 or F=54763822.2/43649115.1. We calculated the

F test statistic to be 1.2546. After finding the test statistic, we proceeded to find the critical

values. First we found that the degrees of freedom for the numerator were N1-1, or 11 and the


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degrees of freedom for the denominator were N2-1, or 13. We decided to choose a significance

level of 5%, meaning our alpha was .05. We chose this alpha because we wanted to be very

confident in our outcome. Because this was a two-tailed test we used =.025. We used an F

table to find our critical values with our right critical value being: F/2,V1,V2and our left critical

value being: 1/F/2,V2,V1. These values, respectively, were 3.2 and 0.295. If our F statistic was

greater than 3.2 or less than 0.295, it would lie in the rejection region and we could reject the

null. 1.25 lies between 3.2 and 0.295 and therefore we concluded that we did not have enough

statistical evidence at a 5% significance level to reject the null hypothesis in favor of the

alternative hypothesis. In other words, the population variances were assumed to be equal.

Through the use of the F-Test, we determined the population variances to be equal thus

indicating that we will use the equal variance t-test of12 formula in order to calculate the t-

test statistic. The independent sample data we collected can be used to calculate the necessary

variables in the formula. Our first set of sample data, the Pac12, provides us with n1= 12 (sample

size/number of schools), x-bar1=$30,356 (sample mean/average cost of schools), and s^2,1=

43649115.1 (sample variance/variance of the costs). The second set of sample data, the SEC,

provides us with n2=14, x-bar2=$23,997, and s^2,2=54763822.2. In addition, the formula calls

for a pooled variance estimator in order to make the estimate more accurate. S^2 p = ((12-

1)*43,649,115.10 + (14-1)*54,763,822.20) / (12+14-2) = 48,743,355.84. The t-test statistic can

therefore be calculated by:

((30,356-23,997)(0-0)) / ( )

The value of this t-test statistic is 2.31534. Also, v, the degrees of freedom, is equal to 24 by 14 +

12 -2.


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In order to interpret the t-test statistic, we must define the rejection region. First,

we determined that we would use a 95% confidence interval to insure the economic significance

and accuracy of our experiment. The test is a one-tailed test because we are determining if the

difference between population means are greater than a value, 0. We calculated the critical value

of t by using a t-distribution chart. It is important to note from a glance, the distribution of this

data, congruent with the t-distribution, is not extremely nonnormal. The critical value against

which we will test our t-test statistic is t sub a,v or t sub .05, 24 as derived by the t-distribution

chart. The value of this critical value is 1.711. This means our reject region is t > 1.711. By

inserting our t-test statistic of 2.315 we can come to the conclusion of rejecting our null

hypothesis as 2.315 > 1.711. This means that H0: 1 - 2 = 0 is rejected in favor of H1: 1 2 >

0. Therefore, based on the samples, we have enough statistical evidence to conclude at a 95%

confidence interval that population mean of u1 is greater than the population mean of u2.

A test of our experiments legitimacy can be applied by using the p-value of our data. We

calculated the one tailed p-value to be 0.0499885 using an online calculator. Because 0.0499885

< our alpha of 0.05, we can reject the null hypothesis.

The confidence level estimator of the difference between the two population

means with equal population variances must be calculated to determine. The lower and upper

limits of our test. The 95% confidence level estimator of our data is depicted is equal to:

(30,356-23,997) +,- 2.064*

By calculating these values, we found the Upper Confidence Level to be $12,028.16 and the

Lower Confidence Level to be $690.34. Therefore, we are 95% certain that the difference


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between the population means falls within the interval of $690.34 to $12,028.16. The difference

between the sample means for our data is $6,359, which falls into this range.

Conclusion

This hypothesis test was a right tailed test with a test statistic of 2.31534. We determined

our critical value at a 5% significance level to be 1.711. The test statistic was larger than the

critical value, and therefore was in the rejection region. Because of this, we had enough

statistical evidence at a 5% significance level to reject the null hypothesis in favor of the

alternative hypothesis. We can conclude that schools in the west have a higher average tuition

than schools in the southeast. Another way to confirm this conclusion was to calculate the p-

value of this hypothesis test. If the p-value was less than the significance level, we could also

reject the null hypothesis in favor of the alternative hypothesis. Our p-value, calculated online,

turned out to be 0.0499885. This gave us enough evidence to reject the null hypothesis in favor

of the alternative. We then proceeded to construct a confidence interval and concluded that we

are 95% certain that the difference between the two means falls between $690.34 and

$12,028.16. In our sample, the difference between the two means was $6,359.

Once again, we concluded that schools in the west have a higher average tuition than

schools in the southeast. In order to come to this conclusion, we used the sample data of Pac12

schools to represent the West and SEC schools to represent the Southeast. By testing these

population means against one another, we were able to reject our null hypothesis of the

differences between the population means being equal. One weakness in our data was the fact

that our sample consisted of sport-dominated schools, which are typically larger markets and

therefore might not be the best representation of the population as a whole. However, since we


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compared the same types of schools against each other, this might not be a serious weakness. In

addition, we do not know if there are more private schools in one region than the other.

Weighting our sample data based on a private/public school ration might change our data as

private schools tend to be more expensive. We could improve this hypothesis test by using a

larger sample size and more years of data.

the cost of education in western united states compared to southeastern united states

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