Download 9.14 Using JMP Example 9.14.1 (Using JMP When Testing a

9.14 Using JMP
Example 9.14.1 (Using JMP When Testing a Population Mean) Refer to Example 8.11.1, where
we have a data set from a normal population with unknown mean  and unknown standard
deviation  . Test, using JMP, (1) The null hypothesis  = 30 against (a)   30 (b)   30 , and
(c)   30 at the 5% level of significance; (2) The null hypothesis   7 against (a)   7 (b)
  7 , and (c)   7 at the 5% level of significance.
23
28
20
25
29
24
20
36
29
16
26
26
19
27
37
35
35
38
42
41
24
25
30
26
Solution: To test a hypothesis about the mean  and the standard deviation  using JMP,
proceed as follows:
1. 1. From the main JMP taskbar, select File > New > Data Table, which results in an untitled
data table in a separate screen. To assign a title to this table, put the cursor in the box titled
“Untitled” in the left corner, right click twice, and enter the desired title.
2. In New Data Table enter the data in Column 1 and label it Data or Sample.
3. Select from the bar menu Analyze > Distribution. In the new dialog box that appears, select
the sample as Y variable. Then click OK.
4. A dialog giving some general results about the distribution appears. From this dialog select the
red triangle next to the data name. From the pull-down menu (shown in Figure 8.11.1) select
Test Mean. Another dialog box appears where we indicate the hypothesized value, say 30 in this
example. Also, if the population standard deviation is known and/or the sample size is large so
that we can use the z-test instead of the t-test, then enter the true value or the estimated value of
the population standard deviation in the box next to the hypothesized value location.
The test result shown in Table 9.14.1 will appear. Note that the test results shown in Table 9.14.1
give the p-value for all three alternative hypotheses. The first is the two-tailed test, the second
and third are for the alternatives of  > and  < the hypothesized value, respectively. All three
p-values are greater than the level of significance 5%. Hence we will not reject the null
hypothesis in any case. The diagram in Table 9.14.1 is the distribution of X under the null
hypothesis.
Table 9.14.1 JMP output for testing the hypothesis
about the mean with unknown variance.
(2) To test the hypothesis about the standard deviation, follow all the steps of part (1), but
instead of selecting Test Mean from the menu select Test Std Dev. The test results in the form
of p-values (last three rows) are shown in Table 9.14.2. By examining the p-values we can
conclude that we do not reject the null hypothesis in any case.
Table 9.14.2 JMP output for testing the hypothesis
about the standard deviation.
Example 9.14.2 ((Using JMP When Testing the Equality of Two Population Means) Refer to
Example 8.11.2, where we are given the total cholesterol level among women by age group.
Assume that the total cholesterol levels of these two segments of the population are normally
distributed with unknown means 1 and  2 and unknown variances  12 and  22 , respectively.
Test, using JMP, the null hypothesis 2  1 = 0 against (a) 2  1  0 , (b) 2  1  0 , and (c)
2  1  0 at the 5% level of significance.
Solution: The solution for this problem proceeds as in Example 8.11.2, so we just reproduce
here the results from Example 8.11.2 and utilize them to answer (a), (b) and (c).
Table 9.14.3 JMP output for confidence interval and testing
of hypothesis for two population means with equal variances
Table 9.14.4 JMP output for confidence interval and testing
of hypothesis for two population means with unequal variances
First, we look at the results in Table 9.14.3 where we assumed that the two population variances
are equal. By examining the middle part of the table, we find that the p-values for the two-tail
and the right-tail tests are less than 0.0001. Thus, in (a) and (b) we reject the null hypothesis in
favor of the alternative hypotheses. However, in (c) the p-value is 1, which implies that in this
case we do not reject the null hypothesis. Combining the conclusions we may state: based on this
data, we can conclude at any significance level greater than 0.0001 that the cholesterol level of
the women in age group 55  70 is higher than those in age group 40  55.
The third portion of Table 9.14.3 gives results for the hypothesized distribution of X 1  X 2 under
H0 .
Figure 9.14.1 Pull-down menu showing JMP’s various options for two populations.
The interpretation of the results in Table 9.14.4 for the case when two population variances not
equal is exactly the same. The power, for example, at   2  1  2 and 9.5 is determined by
selecting the option Power in the pull-down menu shown in Figure 9.14.1 and then entering the
values of  in the dialog box. The power for the test at   2  1  2 and 9.5 is shown in Table
9.14.5, when  is assumed to be 5.586691 (pooled estimate of  , i.e., when two variances are
unknown but assumed to be equal).
Table 9.14.5 JMP output for power of the test at   2  1  2, 9.5.
Testing the variances:
To test the hypothesis  1   2 against  1   2 , follow the same steps as given in Example 8.11.2
and then select from the pull-down menu in Figure 9.14.1 the option Unequal Variances. The
result for the test appears on the screen as shown below.
Note that the JMP output shows results using various tests. However, here we only give the
result obtained by using the F-test.