Math 256, Hallstone Minitab Homework 4 Spring 2008
Inference on Minitab
Redo the 2-proportion problem we did in class with Minitab: A manufacturer experiments with two production methods. The first method produces 18 defects out of 275 sample items, while the second method produces 27 defects out of 320 samples. At the a = .05 level, test the claim that there is no difference between the proportions of defects.
1) Pull down the “Stat" menu.
2) Choose the "Basic statistics” option.
3) Choose “2 Proportions…”
3) Put the bullet in “Summarized Data”, fill in the n values in “Trials”, the y-values in “Events”
4) Click on “Options…”. Under “Confidence level:” use 1- α (in this case .95), put 0 in “Test difference:” and pick the type of alternative. Check the box for “Use pooled estimate of p for test” and hit “OK.”.
5) Again click “OK.”
6) Print the results from the session window.
Note: You could do a clustered bar chart of this using the directions from the earlier handout. I have repeated them here adapted to this example.
Clustered Bar Chart
0) Put the data in a Minitab worksheet. It should look like this:
1) Pull down “Graph” menu.
2) Choose the “Bar Chart…” option.
3) In the “Bars Represent:” box, it should be changed to “values from a table.” Below that under “Two-way table,” highlight “Cluster” and hit “OK.”
4) For the “Graph Variables:,” double click on “Method 1” and “Method 2.”
5) Click on the “Row labels” box, and then double click on “Type of Part”
6) Make sure the bullet is in the choice “Columns are outermost…”, and then click on “Bar Chart Options.” Make sure that “Show Y as Percent” is picked. The last choice at the bottom, “Within categories at level 1 (outermost),” should be bulleted.
7) Click “OK” and then click on “Labels…”. Click on the “Data Labels’ tab, and then bullet the “use y-value labels” option.
8) Click “OK” twice and a window should open with the clustered bar chart. It should look like the following. Note that the numbers are (in percent form) etc.
Redo the 2-mean problem we did in class with Minitab: An education professor at a large university was interested in trying a new method of instruction that involves more student participation than the classical lecture method. To compare the two methods of instruction, she taught two sections of the same course. The mean final grades in each of the two sections were as follows: The 35 students in the lecture method had a mean of 72.3 and a standard deviation of 9.1. The 44 students in the new method had a mean of 74.6 and a standard deviation of 8.8. Test, at the 0.05 significance level, whether or not the methods are equally effective.
1) Pull down the “Stat" menu.
2) Choose the "Basic statistics” option.
3) Choose “2 Sample t…”
3) Put the bullet in “Summarized Data”, fill in the appropriate values for the sample sizes, sample means, and sample standard deviations. Do not check the box for equal variances.
4) Click on “Options…”. Under “Confidence level:” use 1- α (in this case .95), put 0 in “Test difference:” and pick the type of alternative. Click OK.
5) Again click “OK.”
6) Print the results from the session window.
The session window output should look like this:
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 35 72.30 9.10 1.5
2 44 74.60 8.80 1.3
Difference = mu (1) - mu (2)
Estimate for difference: -2.30000
95% CI for difference: (-6.35021, 1.75021)
T-Test of difference = 0 (vs not =): T-Value = -1.13 P-Value = 0.261 DF = 71
Note: If you had the actual data, you could do side-by-side boxplots. With just the summary data, there is really nothing to graph.
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