Question

The nonprofit group Public Agenda conducted telephone interviews with a stratified sample of parents of high school children. There were 202 black parent, 202 Hispanic parents, and 201 white parents. ONe question asked was, “Are the high schools in your state doing an excellent, good, fair, or poor job, or you don’t know enough to say?” Here are the survey results:

	Black	Hispanic	White	Total
Excellent	12	34	22	68
Good	69	55	81	205
Fair	75	61	60	196
Poor	24	24	24	72
Don't Know	22	28	14	64
Total	202	202	201	605

(a) Give a 95% confidence interval for the proportion of families who think that the high schools are doing a “poor” job.

(b) You are interested in determining if the ratings of the schools differ among the sets of parents. Would you use a Chi-Square test of Independence or Homogeneity? Justify your answer.

(c) Use Crunchit to perform the Chi-Square test of significance. Report your findings.

Please Show all work.

Answer 1

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   72          
Sample Size,   n =    605          
                  
Sample Proportion ,    p̂ = x/n =    0.1190          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.013164          
margin of error , E = Z*SE =    1.960   *   0.01316   =   0.0258
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.11901   -   0.02580   =   0.09321
Interval Upper Limit = p̂ + E =   0.11901   +   0.02580   =   0.14481
                  
95%   confidence interval is (   0.093   < p <    0.145   )

b)

Chi-Square Test of independence
	Observed Frequencies
	0
0	Black	Hispanic	White				Total
Excellent	12	34	22				68
Good	69	55	81				205
Fair	75	61	60				196
Poor	24	24	24				72
Don't Know	22	28	14				64
Total	202	202	201				605
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	Black	Hispanic	White				Total
Excellent	202*68/605=22.704	202*68/605=22.704	201*68/605=22.592				68
Good	202*205/605=68.446	202*205/605=68.446	201*205/605=68.107				205
Fair	202*196/605=65.441	202*196/605=65.441	201*196/605=65.117				196
Poor	202*72/605=24.04	202*72/605=24.04	201*72/605=23.921				72
Don't Know	202*64/605=21.369	202*64/605=21.369	201*64/605=21.263				64
Total	202	202	201				605
	(fo-fe)^2/fe
Excellent	5.0466	5.6200	0.015
Good	0.0045	2.6415	2.441
Fair	1.3962	0.3014	0.4022
Poor	0.0001	0.0001	0.0003
Don't Know	0.0187	2.0580	2.4808

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   22.426  
      
Level of Significance =   0.05  
Number of Rows =   5  
Number of Columns =   3  
Degrees of Freedom=(#row - 1)(#column -1) = (5- 1 ) * ( 3- 1 ) =   8  
      
p-Value =   0.0041847   [Excel function: =CHISQ.DIST.RT(χ²,df) ]
Decision:    p-value < α , Reject Ho  
there is enough evidence that the ratings of the schools differ among the sets of parent