There are four entrances to the Government Center Building in downtown Philadelphia. The building maintenance superviso
Entrance Frequency
Main Street 140
Broad Street 120
Cherry Street 90
Walnut Street 50
Total 400
1. State the null and alternate hypothesis
Let H0 = There is no difference in the use of the four entrances
and Ha = There is difference in the use of the four entrances
a = 0.01
2. State the decision rule
Critical value of the chi-square corresponding to dof = 4 - 1 = 3 and a = 0.01 is 11.3449
Reject H0 if the chi-square value obtained > 11.3449
3. Compute the value of the test statistic.
Entrance Frequency Frequency (Recorded - (Recorded -
Recorded Expected Expected)^2 Expected)^2 /
Expected
Main Street 140 100 1600 16.00
Broad Street 120 100 400 4.00
Cherry Street 90 100 100 1.00
Walnut Street 50 100 2500 25.00
Chi-square = 46.00
4. What is the decision. Interpret the results.
Since the test value of the chi-square (46) > the critical value (11.3449), we reject H0 and accept Ha.
It appears that there is difference in the use of the four entrances.
he building maintenance supervisor would like to know if the entrances are equally utilized. T investigates, 400 people were observed en
Entrance F (Recorded) F (Expected)
MS 140 100
BS 120 100
CS 90 100
WS 50 100
Goodness of Fit Test
observed expected O - E (O - E)² / E % of chisq
140 100.000 40.000 16.000 34.78
120 100.000 20.000 4.000 8.70
90 100.000 -10.000 1.000 2.17
50 100.000 -50.000 25.000 54.35
400 400.000 0.000 46.000 100.00
46.00 chi-square
3 df
5.67E-10 p-value
0 people were observed entering the building. The number using each entrance is reported below. At the .01 significance level, is there
he .01 significance level, is there a difference in the use of the four entrances?
Goodness of Fit Test
observed expected O-E (O - E)² / E % of chisq
140 100.000 40.000 16.000 34.78
120 100.000 20.000 4.000 8.70
90 100.000 -10.000 1.000 2.17
50 100.000 -50.000 25.000 54.35
400 400.000 0.000 46.000 100.00
46.00 chi-square
3 df
5.67E-10 p-value