Question 8.1: Purchase Value of Grocery Baskets Study The Grocery Retailer......

This hypothesis test is a two-tailed test for a single mean because the GRASA’s belief is that the population mean of the numeric random variable ‘value of grocery purchase’ is a single value (i.e. R175) only.

Step 1: Define the null and alternative hypotheses

Given that this is a two-tailed test, H_0 and H_1 are formulated as follows:

H_0 : μ = 175 This represents the management claim to be tested.

H_1 : μ ≠ 175

The null hypothesis will be rejected in favour of the alternative hypothesis if the sample evidence shows that the actual mean value of grocery purchases is either significantly less than or significantly more than the null hypothesised value of R175.

Step 2: Determine the region of acceptance of the null hypothesis

A level of significance is needed to find the critical z-limits between the regions of acceptance and rejection. In this example, α = 0.05 (5% level of significance).

Since this is a two-tailed test, the region of acceptance is defined by two limits: a lower and an upper limit around H_0. These two critical z-limits identify a combined area of α = 0.05 in the two tails of the z-distribution.

The critical z-limits are z–crit = ±1.96 (Table 1, Appendix 1). Thus the region of acceptance for H_0 is −1.96 ≤ z ≤ +1.96.

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

• Accept H_0 if z–stat falls between −1.96 and +1.96.
• Reject H_0 if z–stat falls below −1.96 or above +1.96.

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

The sample test statistic, z–stat, is calculated using sample data that is substituted into Formula 8.1.

Given \bar{x}= = 182.40, \sigma = = 67.5 and n = 360. First calculate the standard error:

Then:    z–stat = \frac{182.4-175}{3.558}=\frac{7.4}{3.558}=2.08

This z–stat tells us how many standard errors the sample mean of R182.40 lies away from the null hypothesised population mean of R175. Here, \bar{x} lies 2.08 standard errors above R175. The further away it lies, in standard error terms, the more likely the null hypothesis will be rejected.

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

This sample test statistic, z-stat, must now be compared to the decision rule for the region of acceptance (Step 2) to decide if it is ‘close enough’ to the null hypothesised population mean.

The sample test statistic z-stat = 2.08 lies outside the region of acceptance of −1.96 ≤ z ≤ +1.96. Refer to Figure 8.6, which shows the sample test statistic (z-stat) in relation to the regions of acceptance and rejection as defined by the z-crit limits.

Step 5: Draw statistical and management conclusions

Statistical Conclusion

Since z-stat lies outside the region of acceptanceof H_0, the sample evidence is not ‘close enough’ to the null hypothesised value of R175 to accept H_0. There is sufficient sample evidence at the 5% level of significance to reject H_0 in favour of H_1. The alternative hypothesis is therefore probably true.

Management Conclusion

It can be concluded, with 95% confidence, that the actual mean value of grocery purchases is not R175. The GRASA’s claim cannot be supported on the basis of the sample evidence presented.