Question

The new management at a bakery claims that workers are now more productive than they were under old management, which is why wages have "generally increased." LetWib be workers i's wage under the old management and let Wia be Worker i's wage after the change. The difference is Di = Wia - Wib. Assume that the Di are a randomsample from a Normal (μ,σ²) distribution.

i) Using the following data on 15 workers, construct an exact 95% confidence interval for μ.
ii) Formally state the null hypothesis that there has been no change in average wages. In particular, what is E(Di) under H0? If you are hired to examine thevalidity of the new management's claim, what is the relevant alternative hypothesis in terms of μ = E(Di)?
iii)Test the null hypothesis from part (ii) against the stated alternative at the 5% and 1% levels.
iv) Obtain the p-value for the test in part (iii).

Answer 1

Consider the following table shows the wage under old and new management:

(i)

To calculate the 95% confidence interval for the mean, use pair t-test and use the Minitab software. The procedure is as below:

1. Open the Minitab software, select the stat option and click on the basic statistics then select t-t paired.

2. After click t-t paired option, a new dialog box will open, select the options as shown below:

3. After filling the summarized data, select the option button a new dialog box will open select the confidence interval as 95% and alternative as Not equal then press OK button to return to the old dialog box and then click OK button to get the result. The result of the t-t pair test is as below:

According to the above output, the confidence interval for mean is . In other words, the mean of difference lies in between the confidence interval.

(ii)

If denote the difference between the mean as:

The null hypothesis for the pair t-test (hypothesis of no difference between the mean of two mixtures) is:

And, the alternative hypothesis, against the null hypothesis is:

Here is the mean of old management and is the mean of wage after change.

If the null hypothesis is that the population mean is equal to the sample mean, then the null hypothesis can be written as, the mean of difference between the means of wage before and after change as follows:

(iii)

According to the result obtained in the part (a), the t-value is and the P- value is 0.488. Thus, at 5% significance level, the P-value is greater than the level of significance , so the hypothesis does not get rejected. Thus, the mean of difference between the means of wage before and after change is 0.

At 1% significance level, the P-value is greater than the level of significance , so the hypothesis does not get rejected. Thus, the mean of difference between the means of wage before and after change is 0.

(iv)

According to the results obtained in the part (a), the P-value for the test is approximately .