Question

8. Teachers want to know which days of the week students are absent in a five-day school week. The student absences for one school week are listed in the table below.    Use a 0.05 significance level to test the claim that student absences fit a uniform distribution.

Monday Tuesday Wednesday Thursday Friday

Number of Absences   16         9      11         10        24

USE A 0.05 SIGNIFICANCE LEVEL TO TEST THE CLAIM:

H0 : All the proportions are the same claimed value. (generic null hypothesis) H1: At least one proportion is different. (generic alternative hypothesis)

Expected Value: E =______ Q30(ROUND TO WHOLE)

Test Statistic: X2=_________ Q31(ROUND TO WHOLE)

P-value: p=___________Q32(ROUND TO 4 DECIMAL PLACES)

Decision/Conclusion: _________________________________Q33(CIRCLE ONE BELOW) Reject H0, There is sufficient evidence to warrant the rejection of the claim.

OR

Fail to Reject H0, There is not sufficient evidence to warrant the rejection of the claim.

Answer 1

Q30 . The expected value is same for all the days (as the absences fit a uniform distribution)

Expected value , E= (16+9+11+10+24) /5 =14

Q31. Test statistic

Calculation of Chi square

	O	E	(O-E)^2/E
	16	14	0.285714
	9	14	1.785714
	11	14	0.642857
	10	14	1.142857
	24	14	7.142857
sum			11

Calculation of Chi square

Thus

11

Q32. degrees of freedom = 5-1= 4

P value = 0.0266

Note : Excel formula for P value "=CHISQ.DIST.RT(11,4)"

Q33. Since P value < 0.05 (significance level)

Reject H0 . There is sufficient evidence to warrant rejection of the claim .