Please show the work in how to do these problems: Ally VA Topic Five Quiz x Log
ID: 3216859 • Letter: P
Question
Please show the work in how to do these problems:
Ally VA Topic Five Quiz x Log In Loud Cloud Syste x CA Research Project B x C O www.webassign. t/web/Student/Assignme Response /last? dep 154358644 Q5 5. -3 points MIntroStat8 9 E.514 XP My Notes Ask Your Teacher A research project based on a study of older adults examined the relationship between physical activity and pet ownership. The data collected included information concerning pet owner characteristics and the type of pet owned. Here are data giving the relationship between pet ownership status and gender. Pet ownership status Gender No pet owners Dog owners Cat owners Female 1026 154 84 Male 923 172 83 Analyze the data Round your x2 to three decimal places and your P-value to four decimal places df P-value 3 Summarize your results and conclusions. The relationship between gender and pet ownership is significant. The relationship between gender and pet ownership is not significant. Submit Answer Save Progress 6. -3 points MIntroStat8 10 E.032. My Notes Ask Your Teacher The index of biotic integrity (IBI) is a measure of the water quality in streams. IBI and land-use measures for a collection of streams in the Ozark 6008 PM 4 4/2/2017 Ask me anythingExplanation / Answer
Question 5
Solution:
Here, we have to use the Chi square test for independence of two categorical variables. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The relationship between gender and pet ownership is not significant.
Alternative hypothesis: Ha: The relationship between gender and pet ownership is significant.
We can also write these hypotheses as below:
Null hypothesis: H0: The two categorical variables gender and pet ownership are independent.
Alternative hypothesis: Ha: The two categorical variables gender and pet ownership are not independent.
We assume the level of significance or alpha value for this test as = 0.05 or 5%.
The formula for test statistic is given as below:
Chi square = [(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Expected frequencies are calculated as below:
E = (row total * column total) / grand total
We are given
Total number of rows = r = 2
Total number of columns = c = 3
Degrees of freedom = (r – 1)*(c – 1)
Degrees of freedom = (2 – 1)*(3 – 1) – 1*2 = 2
Degrees of freedom = 2
The calculation tables for this test are given as below:
Observed Frequencies
Pet Ownership Status
Gender
Non-pet owners
Dog owners
Cat owners
Total
Female
1026
154
84
1264
Male
923
172
83
1178
Total
1949
326
167
2442
Expected Frequencies
Pet Ownership Status
Gender
Non-pet owners
Dog owners
Cat owners
Total
Female
1008.819001
168.7403767
86.44062244
1264
Male
940.1809992
157.2596233
80.55937756
1178
Total
1949
326
167
2442
(O - E)
17.18099918
-14.74037674
-2.440622441
-17.18099918
14.74037674
2.440622441
(O - E)^2/E
0.292606238
1.28765095
0.068910169
0.313967984
1.381656028
0.073940962
Test statistic value Chi square is given as below:
Chi square = 3.4187
Critical value = 5.99
P-value = 0.1810
Alpha value = 0.05
P-value > Alpha value
So, do not reject the null hypothesis that the relationship between gender and pet ownership is not significant.
Conclusion:
The relationship between gender and pet ownership is not significant.
Observed Frequencies
Pet Ownership Status
Gender
Non-pet owners
Dog owners
Cat owners
Total
Female
1026
154
84
1264
Male
923
172
83
1178
Total
1949
326
167
2442