Find answers to your questions faster and easier with IDNLearn.com. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
The chi-square test is 5.02 higher than critical value 3.841 so the two states does not differ with regards of shopping habits, and p value is 0.025.
O = observed value
E = expected value
n = size of sample = 2
df = degree of freedom = n - 1 = 2 - 1
α = significance level = 5% = 0.05
Chi-square is a one of many test statistic, chi-square is a test to see the different from observed value and expected value.
Chi-square formula is,
[tex]\chi^2[/tex] = [tex]\sum\frac{(O_i - E_i)^2}{E_i}[/tex]
Since, in question only give us the observed value, we need to calculate the expected value first.
Expected value conscientious state A can be calculate by this formula
E(ca) = total of state A value × total of conscientious value ÷ grand total
Repeat this process to all observed value.
We use excel to simplified and for precise calculation. Then, we put the obtained value from excel formula to Chi-square formula,
[tex]\chi^2[/tex] = 0.69 + 1.69 + 0.76 + 1.88
= 5.02
a) Critical value for df = 1 and α = 0.05 from chi-square table is 3.841
Since, the result is 5.02 higher than 3.841, it can be concluded that the two states does not differ with regards of shopping habits.
Assumptions for we use is the study group is independent and the each expected value is higher than 5.
b) We use chi-square p-value calculator with P([tex]\chi^2[/tex] > 5.02) and df = 1, α = 0.05. So, we get 0.025.
Thus, two states doesn't differ with regards of shopping habits since chi-square result is higher than critical value, and p value is 0.025.
Learn more about chi-square here:
brainly.com/question/29392431
#SPJ4
