Question 8.6: Mineral water bottle fills Study A machine is set to fill pl......

Step 1: Define the null and alternative hypothesis

H_0 : σ² ≤ 25 The management question resides in H_0.

H_1 : σ² > 25

This is a one-sided upper-tailed test.

Step 2: Determine the region of acceptance of the null hypothesis (use α = 0.05)

The test statistic is the chi-square statistic, χ².

The critical chi -square limit is χ²-crit = χ²(0.05,19) = 30.14 (Table 3, Appendix 1).

In Excel, use CHISQ.INV.RT(0.05,19).

Thus the region of acceptance for H_0 is χ² ≤ 30.14.

The decision rule for accepting or rejecting H_0 is as follows:

• Accept H_0 if χ²-stat ≤ 30.14
• Reject H_0 in favour of H_1 if χ²-stat > 30.14

Step 3: Calculate the sample test statistic (χ²-stat)

χ² – stat = \frac{(20-1)(5.8)^2}{5^2} = 25.57 Use     Formula 8.4

Also, the p-value = 0.1427 (In Excel, use CHISQ.DIST.RT(25.57,19) = 0.1427)

Step 4: Compare the sample test statistic to the area of acceptance

The sample test statistic, lies within the region of acceptance of H_0.

(i.e. χ²-stat (= 25.57) < χ²-crit (= 30.14)). Alternatively, p-value (= 0.1427) > α = 0.05.

Refer to Figure 8.14 which shows the sample test statistic (χ²-stat) in relation to the regions of acceptance and rejection of H_0.

Step 5: Draw statistical and management conclusions

Statistical Conclusion

Since χ²-stat lies within the acceptance region, accept H_0 at the 5% level of significance. There is insufficient sample evidence to reject H_0 in favour of H_1.

Management Conclusion

It can be concluded, with 95% confidence, that the variability of bottle fills is no more than σ = 5 ml. Thus the bottling machine is producing bottled mineral water within the product specification limits of variability.