Question

A Worker asked 76 interns to randomly pick a natural number from 1 to 5.

Numbered Picked:

1=8 2=14 3=24 4=21 5=9

Does the data provide enough evidence at the 1% significance level, to show that the student's version of random selection is not the same as simple random sampling?

Answer 1

Category	Observed Frequency (O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
1	8	0.2	76 * 0.2 = 15.2	(8 - 15.2)²/15.2 = 3.4105
2	14	0.2	76 * 0.2 = 15.2	(14 - 15.2)²/15.2 = 0.0947
3	24	0.2	76 * 0.2 = 15.2	(24 - 15.2)²/15.2 = 5.0947
4	21	0.2	76 * 0.2 = 15.2	(21 - 15.2)²/15.2 = 2.2132
5	9	0.2	76 * 0.2 = 15.2	(9 - 15.2)²/15.2 = 2.5289
Total	76	1.00	76	13.3421

Ho: Proportions are same.

H1: Proportions are different.

Test statistic:

χ² = ∑ ((O-E)²/E) = 13.3421

df = n-1 = 4

Critical value:

χ²α = CHISQ.INV.RT(0.01, 4) = 13.2767

p-value:

p-value = CHISQ.DIST.RT(13.3421, 4) = 0.0097

Decision:

p-value < α, Reject the null hypothesis

There is enough evidence to conclude that the student's version of random selection is not the same as simple random sampling.