Question

You are to use the Top 100 public Colleges 2003 data set for this project.

Virginia has six schools on this list, and North Carolina has 5. A researcher hypothesizes that the colleges in Virginia and North Carolina have high four-year graduation rates. This project addresses these hypothesis by asking the following questions:

Va.- 80% 81% 36% 65% 59% 25%

NC - 65% 25% 31% 31% 34%

(1) Is the average 4-year graduation rate for the Virginia colleges in the top 100 public Universities greater than the average of all the top 100 public colleges, and

(2) Is the average 4-year graduation rate for the North Carolina colleges in the top 100 public Universities greater than the average of all the top 100 public colleges?

SECTION I Perform an appropriate hypothesis test for these two questions using a=.10. Show all important values you use as part of your test, including the null hypothesis, alternate hypothesis, test statistic value, and either the rejection region and critical value or p-value depending on the method you choose to use. State the result of your test using a complete sentence in context of the question.

SECTION II Create a 90% confidence interval for the average 4-year graduation rate for the top 100 public Universities in 2003. Using this confidence interval, what would you conclude about your research questions? Why?

SECTION III Did you find a different result than when you used hypothesis test? If so, why do you think the results were different? If not, which method is better to answering your question? Does one method give you better information than the other?

Answer 1

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AS FOR GIVEN DATA..

Avg 4 year graduation rate for Virginia : (80% + 81% + 36% + 65% + 59% + 25%) / 6=325.16

Avg 4 year graduation rate for North Carolina : (65% + 25% + 31% + 31% + 34%) /5 = 158.8

H₀ : ? = 325
H_A : ? > 325

The output tells us that the average of the n = 6 pieces of ductile iron was 325.16 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 6, is 2.06). The test statistic t* is 1.22, and the P-value is 0.117.

If the set shows significance level ? at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were greater than 1.7109 (determined using statistical software or a t-table):

If the set used the P-value approach to conduct his hypothesis test, he would determine the area under at_{n - 1} = t₂₄ curve and to the right of the test statistic t* = 1.22:

In the output above, Minitab reports that the P-value is 0.117. Since the P-value, 0.117, is greater than ? = 0.05, the output set to reject the null hypothesis. There is insufficient evidence, at the ? = 0.05 level

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