ECF6102 Quantitative Skills for Business- The Population Proportion

• A follow-up study will provide a point estimate of the population proportion of orders that exceed 4,000. The study must provide 90% confidence that the point estimate is within 0.10 of the population proportion. If no previous proportion estimate is available (not even that calculated in (d) above), how large a sample would you recommend for this study?
• It is claimed that the average monthly warehouse distribution cost is more than $65,000. Do the data provide significant support for this claim? Use a 5% significance level and the critical value approach (classical approach)Tested at α = 0.05

The critical z value at 5% significance level is 1.645

Comparing the z-computed and the z-critical values we observe that the z-computed value is greater than the z-critical value thus we reject the null hypothesis and conclude that indeed the average monthly warehouse distribution cost is more than $65,000.

• It is claimed that less than 30% of all mail orders received are for orders less than 4,000. How much evidence do the data provide to support this claim? Use a 0.05 level of significance and the p-value approach to test this claim.

Tested at α = 0.05.

The p-value is 0.4646; this value is greater than 5% significance level leading to non-rejection of the null hypothesis. We therefore conclude that the claim that less than 30% of all mail orders received are for orders less than 4,000 is invalid.

• Imagine that your sample data set only included the first twelve months listed in your original data set. For this new data set, distribution costs still follow the normal distribution. With this reduced data set, test the claim that average monthly warehouse distribution cost is more than $65,000 at the1% significance level. Compare this result to your answer in A(e). Can you suggest any reasons for the variation, if one exists.

The critical z value at 1% significance level is 2.33

Comparing the z-computed and the z-critical values we observe that the z-computed value is less than the z-critical value thus we fail to reject the null hypothesis and conclude that the average monthly warehouse distribution cost is not more than $65,000.

Part B

Using the computer software package Phstat (or any Excel based software) to confirm your answers with the appropriate printout attached for: -

• Solution

Hypothesis tested:

The p-value is 0.4645 (left-tailed), this value is greater than α = 0.05, we therefore fail to reject the null hypothesis and conclude that the claim that less than 30% of all mail orders received are for orders less than 4,000 is invalid

• The hypothesis test with the reduced data set that the mean distribution cost is more than $65,000 in Part A(g) (4 marks)

Solution

Comparing the t-computed and the t-critical values we observe that the t-computed value is less than the t-critical value thus we fail to reject the null hypothesis and conclude that the average monthly warehouse distribution cost is not more than $65,000.

Part C

Using: - the complete data set AND - simple ordinary least squares regression formulae based solutions AND - the regression package on the PhStat software.

• Develop two linear models to explain the distribution costs as a function of their (i) Sales (ii) Number of orders Which model best describes the behaviour of distribution costs? Explain your reasons here (4 marks)

Model 2 (Prediction of costs based on number of orders) is the best model to describe the behavior of distribution costs. This is because R² for this model is 0.8442 while the one for model 1 is 0.7092. This means that a larger proportion of variation in distribution costs is explained in model 2 as compared to model 1 making it the best model.

• Develop a multiple regression analysis to test the distribution costs as a function of both Sales and the Number of orders. Distribution Cost = f (Sales , Number of Orders)

As can be seen from the above results, the two variables are all significant in the model (p-value < 0.05). The model can be seen to have improved by including these variables jointly; this is because the value of R² for this model is 0.8759 (a value greater than for either of the two models previous done). This therefore implies a larger proportion of variation in the dependent variable (distribution costs) compared to either of the first two models is explained by the two independent variables in the final model (model 3).

The variable that appears to most influence costs is the number of orders; this is because the variable (number of orders) has a larger standardized beta value as compared to the other variable (sales). This implies that it (orders) has more influence on the dependent variable (distribution costs) as compared to the sales.

The final regression equation model is given as follows

Part D

1. At the 0.05 level of significance, is there a difference in the variance of the waiting time between the two branches? Use the PhStat software to analyse the problem but give the full hypothesis testing steps in your final presentation. (10 marks)

Solution

 (Null hypothesis, variances are equal)

 (Alternative hypothesis, variances are not equal)

From the above table, we can see that the p-value is 0.1900 (a value greater than α = 0.05). We therefore fail to reject the null hypothesis and conclude that the variance of the waiting time between the two branches is equal.

1. Using the results of (a), which t - test is appropriate for comparing the mean waiting time between the two branches? (4 marks)

Two-Sample Assuming Equal Variances would be appropriate t-test to be used to compare the mean waiting time between the two branches. This is because from (a) above, the variance of the waiting time between the two branches were found to be equal.

1. At the 0.05 level of significance, test whether there is evidence of a difference in the mean waiting time between the two branches using the test selected in (b). Use the PhStat software to analyse the problem but give the full hypothesis testing steps in your final presentation. (10 marks)

From the above table, we can see that the p-value is 0.000 (a value less than α = 0.05). We therefore reject the null hypothesis and conclude that the mean waiting time between the two branches is significantly different.

1. Write a short summary of your findings. (8 marks)

Solution

This report sought to analyze and compare waiting time for two bank branches. An independent samples t-test was done to compare the mean waiting time for the two bank branches. Results showed that the Perth CBD Branch (M = 4.29, SD = 1.64, N = 15) had significant difference in terms of the mean waiting time when compared to the Suburban Branch (M = 7.11, SD = 2.08, N = 15), t (28) = -4.13, p < .05, two-tailed. The difference of 2.83 showed a significant difference.  Essentially results showed that the waiting time at Suburban Branch is much longer than that of the Perth CBD Branch.