Answer:
D. 333.9 and 7.81
Step-by-step explanation:
Hello!
The television network and an online network made two polls asking people their opinion about the way President Obama was handling the Iraq War, they were interested in the approval rating regarding his position on the mater. For this, a test for the proportion of those who approve of President Obama's handling of the war was proposed.
H₀: ρNet=ρOnl
You have two separate, apparently independent populations. The people that answered the Television network poll and the people that answered the online poll.
The parameter of interest is the population proportions of the two populations (Television poll vs online poll) To study the proportions you have two statistical ways you can choose.
One is a test for the difference in population proportions using the approximation to a standard normal distribution (Z).
And the second one is using the Chi-Square Homogeneity test. Remember, this tests objective is to test if the categories of the variable under study (X: opinion of people about the way President Obama was handling the Iraq War, Categorized: Approve, Disapprove.) have the same distribution across the populations of interest (Television poll, Online poll)
(You should reach the same conclusion using either test, the difference lies in the power of both, the approximation to Z is more powerful than the non-parametric test)
Since the question is about a Chi-Square value, the test to chose is the Homogeneity test, wherein the null hypothesis you'll state the proportions of the approval rating are the same in both populations, and that the proportion of disapproval rating is also equal in both populations.
Symbolically:
H₀: ρNet = ρOnl
qNet = qOnl
H₁: At least one of the above equality is not met.
Where:
ρ represents the population proportion of the category "Approve"
q represents the population proportion of the category "Disapprove"
The formula for the Chi-Square statistic is:
χ²=∑∑[tex]\frac{(O_ij-E_ij)^2}{E_ij}[/tex]≈χ²[tex]_{(r-1)(c-1)}[/tex] ∀ i= 1, 2, ..., r and j=1, 2, ..., c
c= total of columns
r= total of rows
i= value in a determined row
j= value in a determined column
The formula for the expected frequencies for this test is:
[tex]e_ij= n_j*\frac{o_i.}{n}[/tex]
The expected frequency for those who approve and answered the online poll:
[tex]E_{12}= O_{.2} * (\frac{O_{1.}}{n}) = 958 * (\frac{724}{2077} ) = 333.939[/tex]
(See attachment for tables of observed and expected frequencies)
It's contribution to the statistic value:
[tex]\frac{(385-333.939)^2}{333.939} = 7.807[/tex]
Just in case you're interested. The complete statistic value is:
χ²=[tex]\frac{(339-390.060)^2}{390.06} + \frac{(385-333.94)^2}{333.94} + \frac{(780-728.94)^2}{728.94} + \frac{(573-624.06)^2}{624.06}[/tex]
χ²= 22.3780
I hope you have a SUPER day!
