Question 8.3: SARS e-Filing Completion Time Study SARS (the South African ......

This hypothesis test is a one-sided lower-tailed test for a single mean because SARS claims that the average e-Filing completion time by all salary-earning taxpayers is less than 45 minutes.

Step 1: Define the null and alternative hypotheses

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

H_0: μ ≥ 45

H_1: μ < 45 This represents SARS’ claim to be tested.

Since SARS claims that the average e-Filing completion time is less than 45 minutes, this strict inequality (<) means that the management question resides in the alternative hypothesis, H_1. The null hypothesis must always contain the equality sign.

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

The level of significance is given as 5% (i.e. α = 0.05). Since this is a one-sided lowertailed test, the 5% level of significance is found only in the lower tail of the sampling distribution for the sample mean. The region of acceptance is defined by a critical lower limit only.

Since the population standard deviation, σ, is unknown, the t-statistic is used to find the critical limit for the region of acceptance. The critical t-limit (t-crit) is that t-value that bounds an area of 5% in the lower tail of the t-distribution.

With degrees of freedom of n − 1 = 11, and α = 0.05, the critical t-limit is t-crit = −1.796 (Table 2, Appendix 1). Thus the region of acceptance for H_0 is t ≥ −1.796.

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

• Accept H_0 if t-stat falls at or above −1.796.
• Reject H_0 if t-stat falls below −1.796.

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

From sample data, compute the sample test statistic, t-stat, using Formula 8.2.

Given \bar{x} = 41.5, s = 9.04 and n = 12. First calculate the (estimated) standard error:

Then       t-stat = \frac{41.5-45}{2.6096}=\frac{-3.5}{2.6096}=-1.341

This t-stat value measures the number of standard errors that the sample mean of 41.5 minutes lies from the null hypothesised population mean of 45 minutes. In this case, \bar{x} = 41.5 minutes lies 1.34 standard error units below the hypothesised mean value of 45 minutes. Is this ‘far enough’ away from the null hypothesised mean of 45 minutes to reject the null hypothesis?

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

This sample test statistic, t-stat, must now be compared to the decision rule (from Step 2) to decide if it is ‘close enough’ to the null hypothesised population mean to accept H_0.

Since t-stat = −1.341, it lies inside the region of acceptance of t ≥ −1.796. Refer to Figure 8.8, which shows the sample test statistic (t-stat) in relation to the regions of acceptance and rejection.

Step 5: Draw statistical and management conclusions

Statistical Conclusion

Since t-stat lies within the region of acceptance, the sample evidence is not convincing enough to reject H_0 in favour of H_1. Thus accept H_0 at the 5% level of significance.

Management Conclusion

It can be concluded, with 95% confidence, that the mean e-Filing completion time is not less than 45 minutes. Thus SARS’ claim cannot be supported.