Please try to include the following three elements for each of the question:
? Hypotheses (both null and alternative hypotheses)
? Analysis (Relevant SPSS printouts or Excel)
? Conclusion (interpretation of data)
? Do not need to outline and examine the assumptions that the chosen statistical technique makes (e.g., the distributions have equal variance in a multiple regression analysis) UNLESS stated in question/problem)

1. SPSS computation and Interpretation: The Asian economy faltered during the last few months of 1997. Investors anticipated that the downturn in the Asian economy would have a negative effect on the earnings of companies in the United States during the fourth quarter of 1997. The following sample data show the earnings per share for the fourth quarter of 1996 and the fourth quarter of 1997 (The Wall Street Journal, January 28, 1998).

a. Formulate H0 and H1 such that rejection of H0 leads to the conclusion that the mean earnings per share for the fourth quarter of 1997 are less than for the fourth quarter of 1996.

Null hypothesis (H0) = mean earnings for 1997 are greater than or equal to the mean earnings for 1996

Alternative hypothesis (H1) = mean earnings for 1997 are less than the mean earnings for 1996

Notice that this is a one-sided test because we're testing if one mean is less than another, instead of if the two means are not equal to each other.

b. Use the data in the following table to conduct the hypothesis test. At alpha = 0.10, what is your conclusion?

Company Earnings 1996 Earnings 1997
Atlantic Richfield 1.16 1.17
Balchem Corp. 0.16 0.13
Black & Decker Corp. 0.97 1.02
Dial Corp. 0.18 0.23
DSC Communications 0.15 -0.32
Eastman Chemical 0.77 0.36
Excel Communications 0.28 -0.14
Federal Signal 0.40 0.29
Ford Motor Company 0.97 1.45
GTE Corp 0.81 0.73
ITT Industries 0.59 0.60
Kimberly-Clark 0.61 -0.27
Minnesota Mining & Mfr. 0.91 0.89
Procter & Gamble 0.63 0.71

We already decided that this would be a one-sided test. Looking at the data, it will be a t-test (because we don't know the population standard deviations) and it will be a matched-pairs test (paired-samples test), because each row of the data comes from one company - the two samples are NOT independent as they would be in a two-sample t-test.

I put the numerical data into two columns of an SPSS spreadsheet, and called the columns e_1996 and e_1997. I then went to analyze → compare means → paired samples t-test, and selected both variable names to go into the "paired variables" box. The results of the test are below:

The mean difference between the 1996 and 1997 earnings is 0.1243 with a 95% confidence interval of the difference of (-0.0642, 0.3128). The two-sided p-value is p = 0.178, which gives us a one-sided p-value of p = 0.089. This is low enough to reject the null hypothesis at the 0.10 level. Therefore we can reject the null hypothesis and assume that the earnings in 1997 were less than the earnings is 1996.

2. SPSS computation and Interpretation:

A factorial experiment involving two levels of factor A and three levels of factor B resulted in the following data.

Factor B
Level 1 Level 2 Level 3

Factor A Level 1 135 90 75
165 66 93
Level 2 125 127 120
95 105 136

a. Formulate the appropriate pairs of H0 and H1 (There are more than one pair of hypotheses in this factorial experiment problem)

Main effect hypotheses:

Null hypothesis 1 (H0) = Factor A has no effect on the results; the means will not differ when using level 1 and level 2

Alternative hypothesis 1 (H1) = Factor A has an effect on the results; the means will differ when using level 1 and level 2

Null hypothesis 2 (H0) = Factor B has no effect on the results; the means will not differ when using level 1, level 2, and level 3

Alternative hypothesis 2 (H1) = Factor B has an effect on the results; the means will differ when using level 1, level 2, and level 3

Interaction effect hypotheses:

Null hypothesis 3 (H0) = Factors A and B affect the results the same way, no matter what the level of the other factor

Alternative hypothesis 3 (H1) = The effects of Factors A and B vary depending on the level of the other factor

b. Construct the ANOVA table - no manual calculation and/or "interpretation" are required for 2(b). Please help show me the manual calculation in addition to computer-generated ANOVA table.

To make the ANOVA table, first you have to put the data in SPSS the right way. Put all the numerical data in one column, then make two additional columns - one for Factor A and one for Factor B. The first few rows of data should look like this:

Result FactorA FactorB
135 1 1
165 1 1
90 1 2
66 1 2
... ... ...

To test main effects for Factor A, go to analyze → compare means → one way ANOVA. The dependent list is "result" and the factor is Factor A. The output looks like this:

Because the p-value is p = 0.423, you can't reject the null hypothesis that the results vary with the level of Factor A.

To make the ANOVA table for Factor B, do the same thing, only put Factor B in the box called factor. You should get this:

Again, you can't reject the null hypothesis that the results vary with the level of Factor B.

Making an ANOVA table by hand isn't fun. But here are the steps:

First, the sums of squares. The "treatment sum of squares" or "between groups sum of squares" measures the variation between groups. You calculate it as

where I is the number of groups, ni is the number in a group, Y-bar is the mean of all the data, and Y-bar with a subscript is the mean of the group corresponding to that subscript.

The "error sum of squares" or "within groups sum of squares" measures the variation within groups, presumably caused by individual variation. You calculate it as

the calculation is similar to SSB, but notice the Yij. These are the individual values, corresponding to the jth term in group i.

The total sum of squares is SSB + SSW. It is calculated as

Now, let's look at the degrees of freedom. The df for the total is the sample size minus 1 (df = n - 1). The df for the "between groups" row is the number of groups minus 1 (df = I - 1). The df for the "within groups" row is the sample size minus the number of groups (df = n - I).

The column in the ANOVA table called "mean squares" is simply the sum of squares for each row divided by the df for that row:

MSB = SSB/(I - 1)
MSW = SSW/(n - I)

To calculate the F statistic, just divide the two values in the mean squares column: F = MSB/MSW. Then look at an F distribution with df = I -1, n - I to get the p-value.

c. Test for the interaction effect as well as main effects (no hypotheses necessary, should be in 2a). Use alpha = .05.

To test for the interaction effect, go to analyze → general linear model and put results in the dependent variable box and the factors in the fixed factors. The output looks like this:

The p-values are in the last column. Look at the row for Factor A*Factor B. The p-value is p = 0.022, which is significant at the 0.05 level. Therefore we can reject the null hypothesis and assume that the effects of each factor vary with the level of the other factor (i.e. their effects are not independent of one another).

3. SPSS computation and Interpretation: Barron's conducts an annual review of online brokers, including both brokers that can be accessed via a Web browser, as well as direct-access brokers that connect customers directly with the broker's network server. Each broker's offerings and performance are evaluated in six areas, using a point value of 0-5 in each category. The results are weighted to obtain an overall score and a final star rating, ranging from zero to five starts, is assigned to each broker. Trade execution, ease of use, and range of offerings are three of the areas evaluated. A point value of 5 in the trade execution area means the order entry and execution process flowed easily from one step to the next. A value of 5 in the ease of use area means that the site was easy to use and can be tailored to show what the user wants to see. A value of 5 in the range offerings area means that all of the investment transactions can be executed online. The following data show the point values for trade execution, ease of use, range of offerings, and the star rating for a sample of 10 of the online brokers that Barron's evaluated (Barron's, March 10, 2003). Use alpha = .05.

Broker Trade Execution Use Range Rating
Wall St. Access 3.7 4.5 4.8 4.0
E*TRADE (Power) 3.4 3.0 4.2 3.5
E*TRADE (Standard) 2.5 4.0 4.0 3.5
Preferred Trade 4.8 3.7 3.4 3.5
my Track 4.0 3.5 3.2 3.5
TD Waterhouse 3.0 3.0 4.6 3.5
Brown & Co. 2.7 2.5 3.3 3.0
Brokerage America 1.7 3.5 3.1 3.0
Merrill Lynch Direct 2.2 2.7 3.0 2.5
Strong Funds 1.4 3.6 2.5 2.0

(a) Determine the estimated regression equation that can be used to predict the start rating given the point values for execution, ease of use, and range of offerings (keep all three independent variables in your regression model even if you find some independent variable(s) that is/are NOT statistically significant). Briefly interpret answer.

Put the four numerical columns into SPSS. Then go to analyze → regression → linear. The dependent variable is rating, and the other three are independent variables. You get the following results:

The "model summary" table tells us that this model accounts for 88.6% of the variation in the ratings.

The "coefficients" table tells us that the regression equation is:

y = 0.345 + 0.255x1 + 0.132x2 + 0.459x3

where y = rating, x1 = execution, x2 = use, and x3 = range. At the 0.05 level, only the coefficients for execution and range are significantly different than 0.

(b) Use the F test to determine the overall significance of the relationship. What is the conclusion at the .05 level of significance? Need the relevant hypotheses before trying SPSS analysis.

The F-test was done in the ANOVA table above. F = 15.485, and p = 0.003, which is significant at the 0.05 level.

The null hypothesis is that the value of the dependent variable does not vary with the values of the independent variables. The alternative hypothesis is that the value of the dependent variable does vary with the values of the independent variables.

We can reject the null hypothesis and assume that there is a relationship between the values of execution, use, and range and the value of the rating.

(c) Use the t test to determine the significance of each independent variable. What is your conclusion at the .05 level of significance? Need the relevant hypotheses before trying SPSS analysis.

These t-tests are in the coefficients table above. The t-tests test the null hypothesis that the coefficients are equal to 0 against the alternative hypothesis that they are different from 0.

From the table, we can see that p = 0.25 for execution, p = 0.382 for use, and p = 0.010 for range. Therefore, the coefficients for execution and range are significantly different that 0, but the coefficient for use is not. The variable use can probably be left out of the linear model with little effect on its prediction power, since the values of use seem to have no effect on the value for rating.