The value of test statistic for test of independence is 62.5
The Chi-square test of independence checks that two variables are related or not. We have to count for two nominal variables. We also believe that the two variables are unrelated. The test allows us to determine whether or not our idea is plausible.
The sections that follow go over what we need for the test, how to perform it, understanding the results, statistical details, and p-values.
Given,
the table shows the beverage preferences for random samples of teens and adults.
For calculating the test statistics,
First wee need to calculate the expected values of every observed values.
Expected values for teens,
[tex]E(50)=\frac{250*400}{1000}=100\\\\E(100=)\frac{250*400}{1000}=100\\\\E(200)=\frac{400*400}{1000}=160\\\\E(50)=\frac{100*400}{1000}=40[/tex]
Expected values for adults,
[tex]E(200)=250-100=150\\\\E(150)=250-100=150\\\\E(200)=400-160=240\\\\E(50)=100-40=60[/tex]
Now, the test statistic is given by:
[tex]X^2=\frac{\sum\limits^n_{i=1}(O_i-E_i)^2}{E_i}[/tex]
Where, Ei=expected value and Oi = observed value
[tex]X^2=\frac{(50-100)^2}{100}+\frac{(100-100)^2}{100}+...\frac{(200-240)^2}{240}+\frac{(50-60)^2}{60}=62.5[/tex]
So, the value of test statistic for this test independence is 62.5.
To learn more about test statistic refer here
https://brainly.com/question/29392431
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Your question is incomplete, here is the complete question
The table below gives beverage preferences for random samples of teens and adults.
teens adults total
coffee 50 200 250
tea 100 150 250
soft drink 200 200 400
other 50 50 100
400 600 1000
we are asked to test for independence between age (i.e., adult and teen) and drink preferences. the test statistic for this test of independence is ?
