Question

Directions: Use the Chi-Square option in the Nonparametric Tests menu to answer the questions based on the following scenario. (Assume a level of significance of .05 and use information from the scenario to determine the expected frequencies for each category) During the analysis of the district data, it was determined that one high school had substantially higher Graduate Exit Exam scores than the state average and the averages of high schools in the surrounding districts. To better understand possible reasons for this difference, the superintendent conducted several analyses. One analysis examined the population of students who completed the exam. Specifically, the superintendent wanted to know if the distribution of special education, regular education, and gifted/talented test takers from the local high school differed from the statewide distribution. The obtained data are provided below.

Number of students from the local high school who took the graduate exit exam - Special education - 16; Regular education - 90; Gifted/talented 16

Percent of test taking students statewide who took the Graduate Exit Exam - Special education 8%; Regular Education 79%; Gifted/Talented 13%

1. If the student distribution for the local high school did not differ from the state, what would be the expected percentage of students in each category?

2. What were the actual percentages of local high school students in each category? (Report final answer to two decimal places)

3. State an appropriate null hypothesis for this analysis.

4. What is the value of the chi-square statistic?

5. What are the reported degrees of freedom?

6. What is the reported level of significance?

7. Based on the results of the one-sample chi-square test, was the population of test taking students at the local high school statistically significantly different from the statewide population?

8. Present the results as they might appear in an article.

 

Answer 1

1.Following table shows the expected frequencies for each category:

         

	O	p	E=125*p
special	16	0.08	10
regular	90	0.79	98.75
gifted/talented	16	0.13	16.25

Total of O=122

Total of E=125

2.

Following table shows the observed percentages:

O	Percentage,(O/125)*100
16	12.8
90	72
16	12.8

3.The null hypothesis is:

H0: The student distribution for the local high school did not differ from the state.

4.

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
16	10	3.6
90	98.75	0.77
16	16.25	0.0038

So test statistics is:

5.

Degree of freedom:

df = 3-1=2

6:The level of signficance : 0.05

7.The p-value is 1

Since p-value is greater than 0.05 so we fail to reject the null hypothesis.

We cannot conclude that the population of test taking students at the local high school was statistically significantly different from the statewide population.