In: Statistics and Probability
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.
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 pi 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.