A pit boss is concerned that a pair of dice being used in a cr...

Question

A pit boss is concerned that a pair of dice being used in a craps game is not fair. The distribution of the expected sum of two fair dice is as follows:

Table showing sums of two dice from 2 to 12 with their probabilities ranging from 1/36 to 6/36 for fairness analysis.

The pit boss rolls the dice 400 times and records the sum of the dice. The table shows the results. Do you think the dice are fair? Use the α = 0.01 level of significance.

Table showing frequencies of sums from 400 dice rolls, listing sums 2 to 12 with corresponding observed counts.

Accepted Answer

Welcome back, everyone. In this problem, a baker is concerned about the consistency of their new cookie recipe. The ideal distribution of the number of chocolate chips per cookie is as follows in the table shown. Now the baker bakes 200 cookies and requires the number of chocolate chips each. The table below shows the results. Do you think the cookie recipe is consistent? Use the 0.05 level of significance. A says yes, the cookie recipe appears to be consistent with the ideal proportions, and B says no. The cookie cookie recipe, sorry, appears to be inconsistent with the ideal proportions. Now this problem is a goodness of fit test using the chi squared distribution because we are testing if the observed frequencies match the expected distribution. So let's first state our hypothesis, OK? So first, let's let the null hypothesis be that the distribution. OK. Of the number of chocolate chips per cookie. OK. Is consistent. OK. With The ideal probability distribution. In other words, that the observed frequencies equally expected frequencies. This must mean then that our alternative hypothesis, and if I could uh borrow our statement here just to save some time in writing. Or or alternative hypothesis then. Would be that the distribution of the number of chocolate chips per cookie is not. Consistent with the ideal probability distribution. If that's the case, Then according to our chi square goodness, sorry, or chi squared goodness of fit test, OK. Then our decision rule. OK. Then it's going to be that if our calculated chi squared value is greater than our critical chi squared value, We are going to reject or null hypothesis. Otherwise we fail to reject or null hypothesis. In other words, if the critical value is greater than the calculated value. So at this point you may realize then that we'll need to calculate or chi squared value, get the critical chi squared value for the significance level and the degrees of freedom, and then compare them. So let's start with our calculated chi squared value and to do that. Recall, OK, recall that the formula for that value or calculated chi square test statistic says that the chi squared value, OK, is going to be the sum of observed values minus expected values squared, OK, divided by the expected values. So at this point, you realize that we will need to figure out uh the expected frequency for each category and compared to our our observed frequency thus to help us figure out a calculated squad value. So let's set up a table, OK? So we're gonna have the number of chips for starters. Let me just put a line here, OK? And then we let me. OK, so we have the number of chips. And then we have the observed value. Let's call that O. We have our expected values, which is gonna be equal to the number of cookies here. OK, where the number of cookies, that is the sample size N multiplied by our ideal probability P. OK. So, we'll be able to calculate those. And then once we have those values, we should be able to use it to figure out the difference between O and E. OK. And then we can square that difference. Yeah, we can have that different squared. And then we can take that, so O minus E squared, OK, and divided by divided by E or expected value. So we know that our distribution goes from 12345678 chips, OK? And our observed values from our table was 8, sorry, 22. OK, 30, 45. 29 35 21. And a 10 Now for our first expected value, if we go back to our table before, remember that the ideal probability for one chip, was 0.05. And the baker bakes 200 cookies, so N equals 200. So our expected value then is going to be 200 multiplied by 0.05, which equals 10. Now we're going to repeat the same idea for the probabilities and N, which is going to always be 200 for the remaining expected frequencies. So for two chips, the ideal probability was 0.1, so the frequency would be 200 multiplied by 0.1 or 20. For 3 chips, it was 0.15. So E is gonna be 30 for 4 chips, it was 0.2, so E will be 40. for 5 chips, it was 0.15, so E would be 30, for 6, it was a 0. Sorry, for 5, my apologies. For 5 it was 0.2 as well, so that was 40. And for 6, it was 0.15. So that was what that would be 30. For a 7, it was 0.1, so that would be 20. And for 8, it was 0.05, so that would be 10. And when we sum all those values up again, the sum would be 200. And since all expected frequencies are greater than or equal than or equal to 5, the chi square test assumptions are met. So now, we can go ahead and find our differences. So in the first one, it would be O minus E 8 minus 10, which is -2. When we square that it's 4, and now when we divide 4 by an expected value of 10, it will be 0.4. Next for 2 chips, we're going to have our difference of 2 square that's 4, and thus our chi square statistic would be 0.2. OK? Next, for 3 chips, the difference is 0. So our squared term is 0 and thus our value is 0. For 4, the difference is 5, square that's 25, and 25 divided by 40 would be 0.625. Next, the difference between 29 and 40 is -11, square that's 121, and divided by 40, that's 3.025. I know you can, I, I'll just go ahead in the interest of time and state the remaining values. You can compare the results you get as well, but the remaining values for our differences, or sorry, for 6 would be 1. No, sorry, 5. 25 and 0.833, approximately. For 7, it would be 11 and 0.05, and for 8, it would be 00 and 0. So now that we have all of these values, the sum of chi squared values, OK, is going to be the sum of all the values in this column, OK? So therefore our calculated test squared statistic is going to be equal to 5.133. So we've completed the first part of our problem. No, we need to determine the critical value. How do we do that? Well, next, OK. 4 degrees of freedom, which equals K minus 1 where K is the number of categories, with 8 categories, the degrees of freedom would be equal to 7. So for 7 degrees of freedom. At a significance level alpha of 0.05, buy a chi squared distribution table or your calculator if it's able to do that. The critical chi squared value. Is going to be equal to 14.067. So now that we have both of these values, we can compare. What do you notice? Well, Notice here that our calculated value. Is less than our critical value. And since it's less than our critical value. Then that means based on our decision rule we fail to reject the null hypothesis. So what does that mean? Well, at the 0.05 level of significance there is insufficient evidence to conclude that the distribution of chocolate chips is different from the ideal distribution. Therefore, the cookie recipe appears to be consistent with the ideal proportions. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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