## Question 852:

1

No answer provided yet.We have counts of people entering different doors and can use the Chi-Square Goodness of Fit test to see if the differences in counts suggest something different than equal usage. The Null Hypothesis for this test is that there is no difference in percentages between entrances. The alternative hypothesis is that at least one entrance has a higher or lower percent usage.

Since there are four entrances, we'd expect a 25% usage for each door. We see a percentage breakdown of .35, .30, .225 and .125. Are these differences more than we'd expect from chance alone?

The Chi-Square test statistic is found by using the formula  [ Observed -Expected ]^2 / Expected. Observed is the raw count and expected is the total count times the expected percent. Since there are 400 people, the expected count is 100. For example, on Walnut street, we'd have the formula = (50-100)^2 / 100 = 25. We repeat this for the other three and add up the values to get 16+4+1+25 = 46, which is the chi-square value.

To evaluate this value we look it up in a Chi-Square table of values using 4-1=3 degrees of freedom. We get the p-value which is less than .0001. Since the p-value is less than the alpha significance level of .01 we reject the Null Hypothesis. We conclude that there is a difference in the usage of the four entrances.

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