Direction: This test covers chapters 12 and IB. Points related to each problem a
ID: 3223212 • Letter: D
Question
Direction: This test covers chapters 12 and IB. Points related to each problem are marked after problem number for a total of 150 points. The director of transportation of a large company is interested in the visage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following: She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05. Referring to Table 12-7, the test will involve _ degree(s) of freedom. KEYWORDS: Chi-square test for difference in proportions, degrees of freedom Referring to Table 12-7, the overall or average proportion of local riders is Easy KEYWORDS: Chi-square test for difference in proportions, propertiesExplanation / Answer
Solution:
Here, we have to use the chi square test for independence of two categorical variables. The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: The two categorical variables types of routes and gender of person are independent from each other.
Alternative hypothesis: Ha: The two categorical variables types of routes and gender of person are not independent.
The test statistic formula for this test is given as below:
Chi square = [(O – E)^2/E]
Where, O is the observed frequency and E is the expected frequency.
Expected frequency is calculated as below:
Expected frequency = E = Row total * Column total / Grand total
We are given,
Level of significance = alpha = = 0.05
Total number of rows = r = 2
Total number of columns = c = 2
Degrees of freedom = (r – 1)*(c – 1) = (2 – 1)*(2 – 1) = 1*1 = 1
The calculations table for this test are given as below:
Observed Frequencies
Column variable
Row variable
Male
Female
Total
Local
27
44
71
Non-Local
33
25
58
Total
60
69
129
Expected Frequencies
Column variable
Row variable
Male
Female
Total
Local
33.02326
37.97674
71
Non-Local
26.97674
31.02326
58
Total
60
69
129
( O - E)
-6.023255814
6.023255814
6.023255814
-6.023255814
(O - E)^2/E
1.098607927
0.955311241
1.344847634
1.169432725
Chi square test statistic = 1.098607927 + 0.955311241 + 1.344847634 + 1.169432725 = 4.5682
Chi square critical value = 3.841459
P-value = 0.032571
= 0.05
P-value <
So, we reject the null hypothesis that the two categorical variables types of routes and gender of person are independent from each other.
We conclude that there is sufficient evidence that the two categorical variables types of routes and gender of person are not independent.
Observed Frequencies
Column variable
Row variable
Male
Female
Total
Local
27
44
71
Non-Local
33
25
58
Total
60
69
129
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.