Question

What motivates employees? The Great Place to Work Institute evaluated nonfinancial factors both globally and in the United States. The results, which indicate the importance rating of each factor, are stored in Table 6. These items were chosen carefully to match each by a common characteristic or factor. At the 0.05 level of significance, is there evidence of a difference in the mean rating between global and U.S. employees? The first few lines of data are shown below:

Factor Global US

Being treated with respect 119 123

Work/life balance 111 112

The type of work that you do 110 111

The quality of people you work with 107 111

The quality of the leadership of the organization 107 112

Base Pay 106 104

What is the null hypothesis for the mean test?

2. Referring to Table 6, what is the upper limit of 90% confidence interval?

3. Referring to Table 6, what is the interpretation of the confidence interval?

		Since the confidence interval does include 0, we do believe there is a difference in the mean ratings.
		Since the confidence interval does include 0, we do not believe there is a difference in the mean ratings.
		Since the confidence interval does not include 0, we do believe there is a difference in the mean ratings.
		Since the confidence interval does not include 0, we do not believe there is a difference in the mean ratings.

Answer 1

mean is given by the formula:-

hence mean of global employees is

110

sample standard deviation of global employees is

4.816

and mean of US employees is

112.1667

sample standard deviation of US employees is

6.112

we need to calculate null hypothesis for the mean test which is given by:-

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are and since F = 0.621, then the null hypothesis of equal variances is not rejected.

Based on the information provided, the significance level is and the degrees of freedom are df = 10.

Hence, it is found that the critical value for this two-tailed test is  

The rejection region for this two-tailed test is

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean   is different than at the 0.05 significance level.

2.

We need to construct the 95% confidence interval for the difference between the population means for the case that the population standard deviations are not known.

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are

The critical value for degrees of freedom is . The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

Since we assume that the population variances are equal, the standard error is computed as follows:

Now, we finally compute the confidence interval:

Therefore, based on the data provided, the 95% confidence interval for the difference between the population means , which indicates that we are 95% confident that the true difference between population means is contained by the interval

hence upper 90% confidence limit is 4.912.

3.

interpretation of confidence limit

Since the confidence interval does include 0, we do not believe there is a difference in the mean ratings.