Question

A busy pharmacy has a BMI weight scale installed inside the pharmacy, which is available to customers from Monday through Friday when the pharmacy is open for business. The manager of the pharmacy wanted to investigate if the number of persons utilizing the BMI scale are the same for each of the 5 days of the week. To investigate this, the manager randomly selected one week and counted the number of customers who used the scale on each of the 5 days during this week. The table below presents the information obtained.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Number of customers	260	199	215	299	286

At the 5% level of significance, should the researcher for the pharmacy reject the null hypothesis that the number of customer who use the BMI scale each of the 5 days of the week is the same?

Conduct a goodness of fit test for this above information, using the steps for hypothesis testing below.                          

1. Null and alternative hypotheses
2. Level of significance
3. Test statistics
4. Critical values and rejection region
5. Process of checking to see whether the test statistic falls in the rejection region
6. Conclusion in words

Answer 1

Solve using Excel:

Day	No. of cusomers	Observed proportion(O)	Expected proportion (E)	(Oi - Ei)2 / Ei
Monday	260	0.207	0.20	0.0002
Tuesday	199	0.158	0.20	0.0088
Wednesday	215	0.171	0.20	0.0043
Thursday	299	0.237	0.20	0.0070
Friday	286	0.227	0.20	0.0037
Total	1259		Test statistic	0.0133
Day	No. of cusomers	Observed proportion(O)	Expected proportion (E)	(Oi - Ei)2 / Ei
Monday	260	=B2/1259	=1/5	=(C2-D2)^2/D2
Tuesday	199	=B3/1259	=1/5	=(C3-D3)^2/D3
Wednesday	215	=B4/1259	=1/5	=(C4-D4)^2/D4
Thursday	299	=B5/1259	=1/5	=(C5-D5)^2/D5
Friday	286	=B6/1259	=1/5	=(C6-D6)^2/D6
Total	=SUM(B2:B6)		Test statistic	=SUM(E2:E4)

a. H₀: p₁ = p₂ = p₃ = p₄ = p₅, Number of customer who use the BMI scale each of the 5 days of the week is the same

H₁: At least one p_i is different, Number of customer who use the BMI scale each of the 5 days of the week is not the same

b. Level of significance = 0.05

c. Test statistic: χ² = 0.013

d. Degrees of freedom: df = k-1 = 5-1 = 4

Left-tail critical value (Using Excel CHISQ.INV(probability,df)) = CHISQ.INV(0.05,4) = 0.711

Right-tail critical value (Using Excel CHISQ.INV.RT(probability,df)) = CHISQ.INV.RT(0.05,4) = 9.488

Rejection Region: Reject H0 if -χ² < 0.711 or χ² > 9.488

e. Since test statistic (0.013) is less than critical value (9.488), we do not reject the null hypothesis.

f. Conclusion: Number of customer who use the BMI scale each of the 5 days of the week is not the same.