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