True false: the problems on these pages can be one or two tailed

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Section 9-1 Testing the Difference Between Two Means: Using the z Test
o lan o o is 6.3 hours, and the population standard deviation of those in the second group found
population standard deviation for those in the first age group found by previous studies
481
EXAMPLE 9-1 Leisure Time
A study using two random samples of 35 people each found that the average amount of
time those in the age group of 26–35 vears spent per week on leisure activities was
39.6 hours, and those in the age group of 46-55 vears spent 35.4 hours. Assume that te
by previous studies was 5.8 hours. At a = 0.05, can it be concluded that there is a
significant difference in the average times each group spends on leisure activities?
SOLUTION
Step 1
State the hypotheses and identify the claim.
Ho: H1 = M2
and
Find the critical values. Since a = 0.05, the critical values are +1.96
Step 2
H1: µ1 # µ2 (claim)
and -1.96.
the
Step 3
Compute the test value.
(X1 - X2) – (µ1 – )
z =
(39.6 – 35.4) - 0
4.2
= 2.90
%3D
1.447
6.32
5.82
V n1
n2
35
35
Step 4
Make the decision. Reject the null hypothesis at a = 0.05 since 2.90 > 1.96.
See Figure 9-3.
(lo)
FIGURE 9-3 Critical and Test Values for Example 9–1
oi d
-1.96
+1.96
+2.90
Summarize the results. There is enough evidence to support the claim that
the means are not equal. That is, the average of the times spent on leisure
activities is different for the groups.
Step 5
The P-values for this test can be determined by using the same procedure shown in
Section 8-2. For example, if the test value for a two-tailed test is 2.90, then the P-value
obtained from Table E is 0.0038. This value is obtained by looking up the area for
z = 2.90, which is 0.9981. Then 0.9981 is subtracted from 1.0000 to get 0.0019, Finally.
this value is doubled to get 0.0038 since the test is two-tailed. If a = 0.05, the decision
would be to reject the null hypothesis, since P-value < a (that is, 0.0038 < 0.05), Note:
The P-value obtained on the TI-84 is 0.0037,
The P-value method for hypothesis testing for this chapter also follows the same for-
mat as stated in Chapter 8. The steps are reviewed here.
State the hypotheses and identify the claim.
Step 1
Step 2 Compute the test value.
Step 3
Find the P-value.
Make the decision.
Step 4
Summarize the results.
Step 5
Transcribed Image Text:Section 9-1 Testing the Difference Between Two Means: Using the z Test o lan o o is 6.3 hours, and the population standard deviation of those in the second group found population standard deviation for those in the first age group found by previous studies 481 EXAMPLE 9-1 Leisure Time A study using two random samples of 35 people each found that the average amount of time those in the age group of 26–35 vears spent per week on leisure activities was 39.6 hours, and those in the age group of 46-55 vears spent 35.4 hours. Assume that te by previous studies was 5.8 hours. At a = 0.05, can it be concluded that there is a significant difference in the average times each group spends on leisure activities? SOLUTION Step 1 State the hypotheses and identify the claim. Ho: H1 = M2 and Find the critical values. Since a = 0.05, the critical values are +1.96 Step 2 H1: µ1 # µ2 (claim) and -1.96. the Step 3 Compute the test value. (X1 - X2) – (µ1 – ) z = (39.6 – 35.4) - 0 4.2 = 2.90 %3D 1.447 6.32 5.82 V n1 n2 35 35 Step 4 Make the decision. Reject the null hypothesis at a = 0.05 since 2.90 > 1.96. See Figure 9-3. (lo) FIGURE 9-3 Critical and Test Values for Example 9–1 oi d -1.96 +1.96 +2.90 Summarize the results. There is enough evidence to support the claim that the means are not equal. That is, the average of the times spent on leisure activities is different for the groups. Step 5 The P-values for this test can be determined by using the same procedure shown in Section 8-2. For example, if the test value for a two-tailed test is 2.90, then the P-value obtained from Table E is 0.0038. This value is obtained by looking up the area for z = 2.90, which is 0.9981. Then 0.9981 is subtracted from 1.0000 to get 0.0019, Finally. this value is doubled to get 0.0038 since the test is two-tailed. If a = 0.05, the decision would be to reject the null hypothesis, since P-value < a (that is, 0.0038 < 0.05), Note: The P-value obtained on the TI-84 is 0.0037, The P-value method for hypothesis testing for this chapter also follows the same for- mat as stated in Chapter 8. The steps are reviewed here. State the hypotheses and identify the claim. Step 1 Step 2 Compute the test value. Step 3 Find the P-value. Make the decision. Step 4 Summarize the results. Step 5
C.
answer
Pages
True false: the problems
on these pages can be one-tail
481-482
or two tailed
Transcribed Image Text:C. answer Pages True false: the problems on these pages can be one-tail 481-482 or two tailed
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