Using SPSS to conduct a Paired-samples t-test - SPSS Help
Problem: In a patent for a process for preparing a dry, free-flowing baking mix, a claim was made that significantly greater volume was produced by type A premix than by type B premix. The volumes of the cakes were measured with the following results:
Recipe |
A |
B |
1 |
29 |
29 |
2 |
32 |
30 |
3 |
25 |
23 |
4 |
37 |
36 |
5 |
39 |
38 |
6 |
26 |
27 |
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The six recipes differed somewhat in the amount of water added, beating time, baking temperature, and baking time. Does the data support the claim made in the patent? Make any relevant comments you think are appropriate. State any assumptions you make. Include results and discussions of a randomize test, a non-parametric test, and a t-test. It is important in your discussion that you show your understanding of the requested statistical tests and the assumptions they are based upon. Compare and contrast the various results of the statistical tests, and include results of other tests that support any of the points you raise in your discussion.
Solution: We have the following data
Recipe |
A |
B |
1 |
29 |
29 |
2 |
32 |
30 |
3 |
25 |
23 |
4 |
37 |
36 |
5 |
39 |
38 |
6 |
26 |
27 |
where the data represents the volume of cakes obtained by using 6 different recipes and two different premixes A and B.
In the patent process, it was claimed that a significantly greater volume was obtained by premix B. We are going to study that claim.
There are several ways we can determine whether the volumes differ significantly.
Parametric Analysis
In this case our aim is to make a statement about the population mean volumes for premix A and B. The claim made in the patent is translated into
\[\begin{array}{cc} & {{H}_{0}}:{{\mu }_{A}}\le {{\mu }_{B}} \\ & {{H}_{A}}:{{\mu }_{A}}>{{\mu }_{B}} \\ \end{array}\]where \({{\mu }_{A}}\) and \({{\mu }_{B}}\) correspond to the population mean for premix A and B, respectively. Due to the nature of the data, this corresponds to a right-tailed, paired t-test.
- Assumptions: In order to apply a paired samples t-test, we both involved population to be normally distributed. If the distributions depart radically from normal distributions, we should use a non-parametric test instead, like the sign test.
In order to apply the test, we need the table with the differences, the average difference and standard deviation of the sample.
Recipe |
A |
B |
Difference |
1 |
29 |
29 |
0 |
2 |
32 |
30 |
2 |
3 |
25 |
23 |
2 |
4 |
37 |
36 |
1 |
5 |
39 |
38 |
1 |
6 |
26 |
27 |
-1 |
Average |
0.833333 | ||
St. Dev |
1.169045 |
The t-statistics is given by
\[t=\frac{\bar{D}-{{\mu }_{D}}}{{{s}_{D}}/\sqrt{n}}=\frac{0.8333-0}{1.169045/\sqrt{6}}=1.746075\]For 5 degrees of freedom, and a significance level of \(\alpha =0.05\), the critical t-value is equal to
\[{{t}_{c}}=2.015\]Since the t-statistics is less than the critical value, we fail to reject the null hypothesis, which means that we don’t have enough evidence to support the claim that the population mean volume for premix A is significantly greater than the mean volume for premix B.
Let's do the analysis in SPSS. The data looks like
Then we go to "Analyze", "Compare Means" and "Paired samples T-test. Then we select the variables "A" and "B" and we click OK. The outputs from SPSS are shown below
The significance of the test is \(p = 0.141\), which means that the difference is not significant.
Non Parametric Analysis
We use nonparametric analysis when either the populations are not normally distributed, or if we want to make a claim about the median of premix A and B. We need to compute the table with the differences represented with a "+" or "-". The sign table is shown below:
Recipe |
A |
B |
Difference |
1 |
29 |
29 |
0 |
2 |
32 |
30 |
+ |
3 |
25 |
23 |
+ |
4 |
37 |
36 |
+ |
5 |
39 |
38 |
+ |
6 |
26 |
27 |
- |
The \(x\) statistics corresponds to the number of the less frequent sign. This means that
\[x=1\]The critical value is 0. Since the statistics is not less than or equal to the critical value, we fail to reject the null hypothesis.
Now we perform the analysis with SPSS. We go to "Analyze", "Non-Parametric tests", "2 related samples", and then we click at "Wilcoxon". The output is shown below.
The significance of the test is \(p = 0.129\), which means that the difference is not significant (same as predicted before).
In other words, we again don’t have enough evidence to support the claim that the population mean volume for premix A is significantly greater than the mean volume for premix B, in this case using a non-parametric approach.
Normality Test
We are going to test the data for normality. We use SPSS for that purpose. We go to "Analyze", "Descriptive Statistics" and "Explore". Then we select the variables "A" and "B". Finally, we go to Plots, and we click at "Normality Plots". The outcome is shown below.
This means that we fail to reject the null hypothesis of normality, since the significance is greater than 0.05. This means that we don’t have enough evidence to claim that the data is not normal.
CONCLUSIONS:
- Both the parametric and non-parametric approach don't provide enough evidence to support the claim the population mean volume for premix A is significantly greater than the mean volume for premix B.
- In order to have a more accurate view, we should a have a bigger sample, to make a more precise statement.
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