Question
In: Statistics and Probability
A poll asked 1054 adults "If purchasing a used car made certain upgrades or features more...
A poll asked 1054 adults "If purchasing a used car made certain upgrades or features more affordable, what would be your preferred luxury upgrade? The results indicated that 8% of the males and 14% of the females answered window tinting. The poll description did not state the sample sizes of males and females. Suppose that both sample sizes were 527 and that 44 of 527 males and 74 of 527 females reported window tinting as their preferred luxury upgrade of choice.
Is there evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.05 level of significance?
a. Let group 1 be the males, and let group 2 be the females. State the null and alternative hypotheses.
b. Calculate the test statistic. X2STAT
c.Determine the critical value.
d. State the conclusion.
e.Find the p-value in (a) and interpret its meaning.
What are the answers if 62 males said they prefer window tinting as a luxury upgrade and 465 did not?
f. The new Test Statistic? X2STAT
g. Determine the new p-value.
h. Interpret the p-value.
i. State the conclusion.
j. An earlier Z-test for the difference between two proportions in part (a) resulted in a test statistic of ZSTAT = −2.93 against critical values of −1.96 and 1.96, with a p-value of 0.003. An earlier Z-test for the difference between two proportions in part (c) resulted in a test statistic of ZSTAT = −1.10, with a p-value of 0.270. Compare the results of (a) through (c) to the results of the Z-tests.
Solutions
Expert Solution
Result:
A poll asked 1054 adults "If purchasing a used car made certain upgrades or features more affordable, what would be your preferred luxury upgrade? The results indicated that 8% of the males and 14% of the females answered window tinting. The poll description did not state the sample sizes of males and females. Suppose that both sample sizes were 527 and that 44 of 527 males and 74 of 527 females reported window tinting as their preferred luxury upgrade of choice.
Is there evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.05 level of significance?
a. Let group 1 be the males, and let group 2 be the females. State the null and alternative hypotheses.
H0: There is no association between gender and preferred luxury upgrade
H1: There is an association between gender and preferred luxury upgrade
b. Calculate the test statistic. X2STAT
Calculated the test statistic=8.589
c.Determine the critical value.
Critical value : 3.841
d. State the conclusion.
Since calculated chi square value 8.589 > 3.841, critical value. Ho is rejected.
We conclude that there is an association between gender and preferred luxury upgrade.
e.Find the p-value in (a) and interpret its meaning.
P value 0.0034.
The P-value is the probability of observing a sample statistic as extreme as the test statistic. since p value < 0.05 level, we reject Ho.
|
Chi-Square Test |
||||||
|
Observed Frequencies |
||||||
|
Column variable |
Calculations |
|||||
|
Row variable |
C1 |
C2 |
Total |
fo-fe |
||
|
R1 |
44 |
74 |
118 |
-15 |
15 |
|
|
R2 |
483 |
453 |
936 |
15 |
-15 |
|
|
Total |
527 |
527 |
1054 |
|||
|
Expected Frequencies |
||||||
|
Column variable |
||||||
|
Row variable |
C1 |
C2 |
Total |
(fo-fe)^2/fe |
||
|
R1 |
59 |
59 |
118 |
3.8136 |
3.8136 |
|
|
R2 |
468 |
468 |
936 |
0.4808 |
0.4808 |
|
|
Total |
527 |
527 |
1054 |
|||
|
Data |
||||||
|
Level of Significance |
0.05 |
|||||
|
Number of Rows |
2 |
|||||
|
Number of Columns |
2 |
|||||
|
Degrees of Freedom |
1 |
|||||
|
Results |
||||||
|
Critical Value |
3.8415 |
|||||
|
Chi-Square Test Statistic |
8.5887 |
|||||
|
p-Value |
0.0034 |
|||||
|
Reject the null hypothesis |
What are the answers if 62 males said they prefer window tinting as a luxury upgrade and 465 did not?
f. The new Test Statistic? X2STAT
Calculated the test statistic=1.216
g. Determine the new p-value.
P value = 0.2702
h. Interpret the p-value.
The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since p value > 0.05 level, we fail to reject Ho.
State the conclusion.
Since p value > 0.05 level, Ho is not rejected.
We conclude that there is no association between gender and preferred luxury upgrade.
|
Chi-Square Test |
||||||
|
Observed Frequencies |
||||||
|
Column variable |
Calculations |
|||||
|
Row variable |
C1 |
C2 |
Total |
fo-fe |
||
|
R1 |
62 |
74 |
136 |
-6 |
6 |
|
|
R2 |
465 |
453 |
918 |
6 |
-6 |
|
|
Total |
527 |
527 |
1054 |
|||
|
Expected Frequencies |
||||||
|
Column variable |
||||||
|
Row variable |
C1 |
C2 |
Total |
(fo-fe)^2/fe |
||
|
R1 |
68 |
68 |
136 |
0.5294 |
0.5294 |
|
|
R2 |
459 |
459 |
918 |
0.0784 |
0.0784 |
|
|
Total |
527 |
527 |
1054 |
|||
|
Data |
||||||
|
Level of Significance |
0.05 |
|||||
|
Number of Rows |
2 |
|||||
|
Number of Columns |
2 |
|||||
|
Degrees of Freedom |
1 |
|||||
|
Results |
||||||
|
Critical Value |
3.8415 |
|||||
|
Chi-Square Test Statistic |
1.2157 |
|||||
|
p-Value |
0.2702 |
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