1. Variable Description :

ATM withdrawal: Amount of euro currency withdrawn

For 300 million people in 12 Eurozone countries, the use of ATM machines is making it easier to use the new euro currency. ATM machines are handling up to 90% of the 14.5 billion euro notes issued. Financial analysts have warned that if people have a bad experience with their ATM that could undermine their belief in the currency. Suppose that a particular bank is interested in how much cash to keep in a local ATM machine. The bank manager believes that the average withdrawal from the local ATM machine on the weekend is 150 euros. A random sample of 60 transactions was recorded. Assume a standard deviation of 50 euros.

1. Plot a histogram of the amount of euros withdrawn using the local ATM machine. Describe the shape of the distribution.
   Solution: With Minitab (Graph>Histogram), we get the following histogram
   
   The shape of the distribution looks pretty much bell-shaped (like a normal distribution)
2. Is there sufficient evidence to indicate that the average withdrawal differs from 150 euros? Use a 5% significance level. Would your conclusion change if you used a significance level of 10%?
   Solution: Using Minitab we perform a two-tailed z-test (Stat > Basic Statistics> 1-Sample z test). The results are shown below:
   
   Since the p-value is p = 0.290, which is greater than 0.05, we fail to reject the null hypothesis. This means that we don’t have enough evidence to claim that the average withdrawal differs from 150 euros, at the 0.05 significance level.
   At the 0.10 significance level, the conclusion would be the same (because p = 0.290 > 0.10)
3. Construct a 95% confidence interval for the average withdrawal. Is the confidence interval consistent with the conclusion in part b?

Solution: From the Minitab output, we find that the 95% confidence interval is

This interval is consistent with the conclusion found in (b), because the confidence interval contains the hypothesized mean 150.

2) Variable description :

StressIndex: Stress index for managers

An industrial psychologist has a stress test that is used to determine the amount of stress that managers are under. A value of 80 or higher indicates "high stress." The industrial psychologist believes that the managers at a large, profitable pharmaceutical firm are not under "high stress" and that the average stress index is less than 80 for managers of the company. A random sample of 50 managers is selected, and their stress index was recorded.

1. Test that the data support the industrial psychologist’s belief.
2. What conclusion would you reach if the significance level was .10? .05? .01?

Solution: Using Minitab we perform a left-tailed t-test (Stat > Basic Statistics> 1-Sample t test). The results are shown below:

The t-statistics is t = -2.03, which means that for 49 degrees of freedom the p-value is p = 0.024. Therefore, we reject the null hypothesis at the 0.10 and 0.05 significance level, but NOT at the 0.01 significance level.

3) Variable description

ThisyearSales: Annual sales of retail stores with advertising on cable channels

LastyearSales: Annual sales of retail stores without advertising on cable channels

Cable television has allowed local retail stores to advertise on certain cable channels. A random sample of 40 retail stores that advertised this year but did not advertise last year was selected. The annual sales were recorded in units of thousands of dollars under the variable names ThisyearSales and LastyearSales.

Is there enough evidence to infer at the 5% significance level that the mean sales from this year is greater than the mean sales for last year for retail stores that started advertising on cable this year?

Solution: We need to use a paired-samples t-test with Minitab (Stat > Basic Statistics > Paired t). The test is a right-tailed one. The results are shown below:

4) Variable description:

ReadReport: A 1 indicates that a shareholder reads the annual report, and a 0 indicates that the shareholder does not.

The vice president of a public utilities company believes that less than 60% of the shareholders of the company read the annual report. From a random sample of 200 shareholders, a 1 or a 0 is listed for the variable ReadReport. Test the vice president’s belief. Use a .01 significance level.

5) Variable description:

MalesInterested: A Yes indicates that a male is interested in a sales manager position, and a No indicates no interest.

FemalesInterested: A Yes indicates that a female is interested in a sales manager position, and a No indicates no interest.

The vice president at Global Life Insurance Company is interested in the proportion of male college business students and female college business students that would be willing to interview for a sales manager position at the company. Random samples of 170 males and 115 females from college business students were selected. Responses were recorded as Yes or No and are listed in variables MalesInterested and FemalesInterested. Test that the proportion of male students exceeds the proportion of female students interested in this position. Use the p-value of test statistic as the basis for your conclusion.

Price: $14.48

Solution: The downloadable solution consists of 6 pages, 848 words and 4 charts.
Deliverable: Word Document