Ctr Chi-Square Test for ation 10.7 following data represent the observed values

Question

Ctr Chi-Square Test for ation 10.7 following data represent the observed values and expected values of the fat contents and ds of microwaveable meals sold at a local grocery store for a random sample of meals sold. bran Observed Sample of 3283 Microwaveable Meals Sold Less Than 5 g Fat 160 86 5-10 g Fat More Than 10 g Fat Total Brand A Brand B Brand C Total 1380 789 of 1567 301 43 3283 Expected Values 5-10 g Fat 20.524216 11.578434 10.897350 43 More Than 10 g Fat 1402.806275 91.372525 744.821200 2939 Les 5 g Fat Total 1567 Brand A Brand B Brand C Total 9510 041 81450 3283 301 data renresent the observed values and expected values of the movie preferences

Explanation / Answer

Hello,

Hope you are keeping well.

Since you have not posted anything regarding what you need to find from the tables, I'm assuming that you are confused as to how you are finding the expected values from the original table.

Here, we are using the Chi-Square test for association. We found out the Expected table to be the sum of elements in the row multiplied by the sum of elements in the columns divided by the grand total.

Now to find the Chi-Squared value for Categorical values. Here you can do it with a very simple method.

We already have a table with the "observed" numbers and the "expected" numbers (i.e. our null hypothesis).

Then subtract each "expected" value from the corresponding "observed" value (O-E).

Square the "O-E" values, and divide each by the relevant "expected" value to give (O-E)2/E.

Add all the (O-E)2/E values and call the total "X2"

You will see that the total value will be coming to 13.33592.

Now we must compare our X2 value with a c2 (chi-squared) value in a table of c2 with n-1 degrees of freedom (where n is the number of categories, in our case 3). We have only two degrees of freedom (n-1).

If our calculated value of X2 exceeds the critical value of c2 then we have a significant difference from the expectation. We see from the Chi-Square table that the value is 5.99 for 5% significance level.

Therefore we have no reason to reject the null hypothesis. Hence we can conclude that the goodness of fit is good enough for the data and it is just fine.

Hope it helped. Cheers!

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