Chi square independence test (nationality & age of children)

[thordogg24]

The following samples of children in daycare facilities in Canada and the US who are less than 2.5 years old or greater than that. Test at 10 percent level of confidence the null hypothesis that there is no relationship between nationality and average age of children in daycare.

Less than 2.5 yrs More than 2.5 years
Canada 18 10
United States 17 13

Ho = No relationship between nationality and average age of children
Ha = Relationship between nationality and average age

Not sure what to do with this? Any help with this would be greatly appreciated. Thanks

 

[royhaas]

Start with the expected values: Row Total times Column Total divided by Grand Total.

 

[thordogg24]

Here is what I have so far, does that look right?

X2= 58[(18)(13) – (10)(17)]^2/ (28)(30)(23)(35) = 237,568/ 676,200 = .35132801

X2 = .35132801

Critical Value = 2.71

Significance level = .10

Degrees of Freedom = (r-1)*(c-1)= (2-1)(2-1)= 1


Ho = No relationship between nationality and average age of children
Ha = Relationship between nationality and average age

 

[galactus]

Do you have to do this by hand with those tedious formulas?.

A lot of calculators will do these.

Using Excel, I get a test stat of .351, critical value of 2.71, p-value of .553

Therefore, do not reject H_0

Now, interpret.