MINITAB Assignment 6
(due Friday May 11)

Comparing two population means

< Text (such as this) which is enclosed in angle brackets, < >, is communication about what to do in the assignment, or commentary about it. I will also color such text green. Text which is not green and bracketed is text you should type or paste into the session window, perhaps with obvious changes, such as replacing Your Name with your name. This will make your minitab work more meaningful on review and in execution. Please try to understand the exercise.>

Minitab Assignment 6: Comparing two population means

PROBLEM 1: Independent samples, pooled

In this problem we use Minitab to work problem 10.17 from the text (p.434), which is copied now.
We are given the following sample weights   of eight female and eleven male wolves:

Female: 57 84 90     71 71 77     68 73
    Male: 71 93 101   84 88 117   86 86 93   86 106

(a) Test the null hypothesis that the mean weights of males and females are equal versus a two-sided alternative. Take alpha =.05 .
(b) Obtain a 95% confidence interval for the difference of population mean weights (female - male).
(c) State in the session window any assumptions you make about the populations.

(c) < State any assumptions. >

(a,b) In column C1 of the data window, type in the title Female (in the box below "C1") and type in the given female wolf weights.
Similar give C2 the title Male and type in the male wolf weights.
Apply Stat > Basic Statistics > 2-sample t, and click on samples in different columns.
Enter Male and Female for the first and second column.
Click assume equal variances, OK.
< You will get Test and CI information appearing in your session window.
Remember to answer the question: at this level of significance, do the results justify the claim that on average male and female wolves have different weight? >

Do the given sample standard deviations justify the assumption of approximately equal variance needed for this pooled test?
< Justify your answer, briefly. >

< Remark: the procedure above, but without "assume equal variances", would produce a conservative t-test and confidence interval.
Here Student Minitab 12 uses a formula for the number of degrees of freedom of the t statistic which is more complicated than the simple formula used by our text, but which can produce a larger number of degrees of freedom and thereby a smaller [better] confidence interval. >

PROBLEM 2: Matched pairs

In this problem we use Minitab to work problem 10.39 from the text (p.434), which is copied now.
It is claimed that an industrial safety program is effective in reducing the loss of working hours due to factory accidents. The following data are collected concerning the weekly loss of working hours due to accidents in six plants both before and after the safety program is instituted.

Plant        1      2       3        4      5      6 
Before    12    30    15      37    29    15
After      10     29    16     35    26    16

Do the data substantiate the claim? Test at level of significance alpha = .05 . State any distributional assumption you make to justify the test.
For this, title C3 and C4 Before and After and type the corresponding data into those columns.
Use Stat > Basic Statistics > paired t.
Choose the settings appropriate to the problem and complete the test.

< Don't forget to answer the questions. >
< Hand in your completed session window. >