In: Statistics and Probability
Please assess whether preference for nuclear power is independent of country of origin at the 95% confidence level.
Select one: a. The chi square statistics is far greater than chi critical - we can reject the hypothesis that nuclear power preference is independent of country. b. The sample is too small to reject the null hypothesis. c. The chi square statistic is greater than chi critical, so we cannot reject the null. d. We can reject the null because the F statistic is greater than F critical.
Background Info:
F | |||||||
Response | England | France | Italy | Spain | Germany | US | |
Strongly Favor | 141 | 161 | 298 | 133 | 128 | 204 | 1065 |
Favor more than oppose | 348 | 366 | 309 | 222 | 272 | 326 | 1843 |
Oppose more than favor | 381 | 334 | 219 | 311 | 322 | 316 | 1883 |
Strongly oppose | 217 | 215 | 219 | 443 | 389 | 174 | 1657 |
1087 | 1076 | 1045 | 1109 | 1111 | 1020 | 6448 | |
E | |||||||
Response | England | France | Italy | Spain | Germany | US | |
Strongly Favor | |||||||
Favor more than oppose | |||||||
Oppose more than favor | |||||||
Strongly oppose | |||||||
Error^2/e: | |||||||
Response | England | France | Italy | Spain | Germany | US | |
Strongly Favor | |||||||
Favor more than oppose | |||||||
Oppose more than favor | |||||||
Strongly oppose |
a. The chi square statistics is far greater than chi critical - we can reject the hypothesis that nuclear power preference is independent of country.
Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: The preference for nuclear power is independent of country of origin.
Alternative hypothesis: Ha: The preference for nuclear power is not independent of country of origin.
We are given confidence level = 95% = 0.95, so level of significance = α = 1 – 0.95 = 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 = 4
Number of columns = c = 6
Degrees of freedom = df = (r – 1)*(c – 1) = 3*5 = 15
α = 0.05
Critical value = 24.99579
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
F |
Observed (O) |
||||||
Response |
England |
France |
Italy |
Spain |
Germany |
US |
Total |
Strongly Favor |
141 |
161 |
298 |
133 |
128 |
204 |
1065 |
Favor more than oppose |
348 |
366 |
309 |
222 |
272 |
326 |
1843 |
Oppose more than favor |
381 |
334 |
219 |
311 |
322 |
316 |
1883 |
Strongly oppose |
217 |
215 |
219 |
443 |
389 |
174 |
1657 |
Total |
1087 |
1076 |
1045 |
1109 |
1111 |
1020 |
6448 |
E |
Expected (E) |
||||||
Response |
England |
France |
Italy |
Spain |
Germany |
US |
|
Strongly Favor |
179.5371 |
177.7202 |
172.6 |
183.1708 |
183.5011 |
168.4708 |
|
Favor more than oppose |
310.6918 |
307.5478 |
298.6872 |
316.98 |
317.5516 |
291.5416 |
|
Oppose more than favor |
317.435 |
314.2227 |
305.1698 |
323.8596 |
324.4437 |
297.8691 |
|
Strongly oppose |
279.3361 |
276.5093 |
268.543 |
284.9896 |
285.5036 |
262.1185 |
|
Error^2/e: |
(O - E)^2/E |
||||||
Response |
England |
France |
Italy |
Spain |
Germany |
US |
|
Strongly Favor |
8.27186 |
1.573067 |
91.10747 |
13.74185 |
16.78666 |
7.492816 |
|
Favor more than oppose |
4.479997 |
11.10938 |
0.356072 |
28.45984 |
6.53422 |
4.072777 |
|
Oppose more than favor |
12.72861 |
1.24479 |
24.33149 |
0.510624 |
0.018406 |
1.103603 |
|
Strongly oppose |
13.91079 |
13.6827 |
9.140082 |
87.6077 |
37.51796 |
29.6235 |
Chi square = ∑[(O – E)^2/E] = 425.4063
P-value = 0.00
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is sufficient evidence to conclude that the preference for nuclear power is not independent of country of origin.