11b: A researcher wishes to test whether the proportion of...

11b: A researcher wishes to test whether the proportion of college students who smoke is the same in four different colleges. She randomly selected 100 students from each college and records the number who smoke. The results are summarized in the table below. \begin{tabular}{|l|c|c|c|c|} \hline & College A & College B & College C & College D \\ \hline Smoke & 17 & 26 & 11 & 34 \\ Don't Smoke & 83 & 74 & 89 & 66 \\ \hline \end{tabular} Conduct the test at the 1\% significance level. You can use StatCrunch for this problem, but you still must show the formula that you are using and briefly describe how it works.

Transcript

00:01 Ok, so we've got expected percentages and observed frequencies, and we want to test whether exam grades differs from what it was in the past.

00:16 So if this year's, the null hypothesis is that this year's grades follows the past distribution, and the alternative hypothesis is that it doesn't.

00:31 And to do that we do a chi -square goodness of fit test where we calculate the chi -square test statistic as the sum of the observed values minus the expected values squared over the expected values.

00:44 So here the observed values were given in the question.

00:47 I calculated the expected values by doing 247 which is the sample size times by the expected percentages.

00:58 So for example the first one would be times by 0 .279 for 29 .7 percent.

01:02 The second one would be times 0 .29 and so on and so on.

01:07 And if we calculate the chi -square test statistic for those sets of observed and expected values, we find that we get 6 .698.

01:15 Our degrees of freedom is just one less than the number of categories, so that's 5 minus 1 which is 4.

01:21 And the p -value associated to this chi -square value with 4 degrees of freedom just turns out to be 0 .153.

01:34 We're asked to do the test at a five percent level and so we say that since our p -value is bigger than the significance level of five percent then we fail to reject h naught and say it just follows the old distribution or there's not sufficient evidence to suggest that it doesn't follow the old distribution.

01:58 The requirements in order to do a chi -squared goodness fit test, we need frequencies for a categorical variable...

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