In: Statistics and Probability
| 
 State  | 
 Below Basic  | 
 Basic  | 
 Proficient  | 
 Advanced  | 
| 
 VA  | 
 360  | 
 1200  | 
 1140  | 
 270  | 
| 
 SC  | 
 609  | 
 1276  | 
 870  | 
 145  | 
| 
 NC  | 
 572  | 
 1848  | 
 1628  | 
 352  | 
Use the chi-square test to determine if there is relationship
between state of residence and NAEP performance level? Write a
short summary of your results. Be sure to include the appropriate
statistical evidence: test statistic used, degrees of freedom,
value of the test statistic, and test statistic critical value. Use
α = .05.
Ans:
Chi square test for independence:
| Observed(fo) | |||||
| State | Below Basic | Basic | Proficient | Advanced | Total | 
| VA | 360 | 1200 | 1140 | 270 | 2970 | 
| SC | 609 | 1276 | 870 | 145 | 2900 | 
| NC | 572 | 1848 | 1628 | 352 | 4400 | 
| Total | 1541 | 4324 | 3638 | 767 | 10270 | 
| Expected(fe) | |||||
| State | Below Basic | Basic | Proficient | Advanced | Total | 
| VA | 445.64 | 1250.47 | 1052.08 | 221.81 | 2970 | 
| SC | 435.14 | 1220.99 | 1027.28 | 216.58 | 2900 | 
| NC | 660.21 | 1852.54 | 1558.64 | 328.61 | 4400 | 
| Total | 1541 | 4324 | 3638 | 767 | 10270 | 
| Chi square=(fo-fe)^2/fe | |||||
| State | Below Basic | Basic | Proficient | Advanced | Total | 
| VA | 16.46 | 2.04 | 7.35 | 10.47 | 36.31 | 
| SC | 69.46 | 2.48 | 24.08 | 23.66 | 119.68 | 
| NC | 11.79 | 0.01 | 3.09 | 1.67 | 16.55 | 
| Total | 97.71 | 4.53 | 34.52 | 35.79 | 172.54 | 
Chi square
Test statistic:=172.54
degree of freedom=(3-1)*(4-1)=6
Critical chi square=CHIINV(0.05,6)=12.592
p-value=CHIDIST(172.54,6)=0.0000