Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: Two categorical variables internet use and household income are independent.
Alternative hypothesis: Ha: Two categorical variables internet use and household income are dependent.
We assume level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
We are given
Number of rows = r = 5
Number of columns = c = 3
Degrees of freedom = df = (r – 1)*(c – 1) = 4*2 = 8
α = 0.05
Critical value = 15.50731
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Observed Frequencies |
||||
How often do you use the Internet? |
||||
Household Income |
Every day |
At least once a week but not every day |
Once a week or less |
Total |
Less than $25,000 |
105 |
46 |
21 |
172 |
$25,000 to $49,999 |
147 |
35 |
17 |
199 |
$50,000 to $84,999 |
211 |
27 |
11 |
249 |
$85,000 to $124,999 |
213 |
25 |
7 |
245 |
$125,000 or more |
161 |
17 |
1 |
179 |
Total |
837 |
150 |
57 |
1044 |
Expected Frequencies |
||||
How often do you use the Internet? |
||||
Household Income |
Every day |
At least once a week but not every day |
Once a week or less |
Total |
Less than $25,000 |
137.8966 |
24.71264 |
9.390805 |
172 |
$25,000 to $49,999 |
159.5431 |
28.59195 |
10.86494 |
199 |
$50,000 to $84,999 |
199.6293 |
35.77586 |
13.59483 |
249 |
$85,000 to $124,999 |
196.4224 |
35.20115 |
13.37644 |
245 |
$125,000 or more |
143.5086 |
25.71839 |
9.772989 |
179 |
Total |
837 |
150 |
57 |
1044 |
Calculations |
||
(O - E) |
||
-32.8966 |
21.28736 |
11.6092 |
-12.5431 |
6.408046 |
6.135057 |
11.37069 |
-8.77586 |
-2.59483 |
16.57759 |
-10.2011 |
-6.37644 |
17.49138 |
-8.71839 |
-8.77299 |
(O - E)^2/E |
||
7.84779 |
18.33683 |
14.35164 |
0.986125 |
1.436175 |
3.464255 |
0.647663 |
2.15273 |
0.495271 |
1.399109 |
2.956251 |
3.039595 |
2.131916 |
2.955486 |
7.875311 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 70.07614
χ2 statistic = 70.07614
P-value = 0.0000
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is not sufficient evidence to conclude that two categorical variables internet use and household income are independent.
Part a)
Answer:
The required expected values are given as below:
Expected Frequencies |
||||
How often do you use the Internet? |
||||
Household Income |
Every day |
At least once a week but not every day |
Once a week or less |
Total |
Less than $25,000 |
137.9 |
24.7 |
9.4 |
172 |
$25,000 to $49,999 |
159.5 |
28.6 |
10.9 |
199 |
$50,000 to $84,999 |
199.6 |
35.8 |
13.6 |
249 |
$85,000 to $124,999 |
196.4 |
35.2 |
13.4 |
245 |
$125,000 or more |
143.5 |
25.7 |
9.8 |
179 |
Total |
837 |
150 |
57 |
1044 |
Part b)
Answer:
χ2 statistic = 70.07614
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