Question

(9 points) Education. Post-secondary educational institutions in the United States (trade schools, colleges, universities, etc.) traditionally offer four different types of degrees or certificates. The U.S. Department of Education recorded the highest degree granted by each of these institutions in the year 2003. The percentages are shown in the table below. A random sample of 250 institutions was taken in 2013 and the number of institutions in the sample for each category is also shown in the table. Conduct a hypothesis test to determine whether there has been any change from the percentages reported in 2003. Round all calculated values to four decimal places.

Highest Degree Awarded	Population percentages in 2003	Sample counts in 2013
Certificate	35.0%	89
Associates	26.6%	60
Bachelor's	11.3%	31
Graduate	27.1%	70

a. Enter the expected values for the hypothesis test in the table below.

Highest Degree Awarded	Expected value
Certificate
Associates
Bachelor's
Graduate

b. Calculate the test statistic for this hypothesis test.   ? z t X^2 F  =

c. Calculate the degrees of freedom for this hypothesis test.

d. Calculate the p-value for this hypothesis test. p-value =

e. Based on the p-value, we have:
A. very strong evidence
B. strong evidence
C. some evidence
D. extremely strong evidence
E. little evidence
that the null model is not a good fit for our observed data.

Answer 1

no change in the percentages reported in 2002) Ho: There is HA: There is change in the percentages reported in 2003. n = 250 a. The expected frequencies Ei are calculated using Ei- napi Highest degree awarded Expected frequency, certificate np, = 250X0.35 = 87.5 Associatetes np2 = 250 X 0.266 = 66.5 Bachelor's np3 = 250 xoll3 = 28.25 Graduate nu= 250 X 0.271 = 67.75 b. Highest degree/ observed | Expected awarded I value (0) I value (E;) I (0; -E;)² I Ei Certificate 87.5 0.0257 60 66.5 0,6353 Associates Bachelor's Graduate 28.25 0.2677 0.0747 70 67.75 Total 1.0034 test statistic to Elized = 1.0034

df = Number of Ei-1 - 4- d. p-value = P( R > X) = P(X3> 1.0034) = 0.8004 To find Ti-83 p-value we use Calculator following command on a and vars . ~ 7.8²cdf x²cdf (1.0034,999,3) e. let 2=0.0s P-value = 0.8004 p-value is much greater than a in we fail to Reject Ho- we Accept Ho that there is no change in percentages reported in 2003, the - we have not a little evidence that the hull model is good fit for our observed data.