Win 4 Lottery Shown below is a histogram of digits selected in...

Question

Win 4 Lottery Shown below is a histogram of digits selected in California’s Win 4 lottery. Each drawing involves the random selection (with replacement) of four digits between 0 and 9 inclusive.

c. Identify the frequencies, then test the claim that the digits are selected from a population in which the digits are all equally likely. Is there a problem with the lottery?

Histogram showing frequency of digits 0-9 drawn in California's Win 4 lottery, highlighting uneven distribution.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says shown below is a histogram of the last digit, the unit's place, of scores from a large set of test results from a high school math department. Each score is out of 100 and the last digit 0 through 9, is recorded for analysis. Identify the frequencies of each digit, then test the claim that the last digits of the scores are uniformly distributed. That is, each digit from 0 to 9 is equally likely. Use 5% level of significance. Does the data suggest any potential grading bias or irregularity in how the scores are assigned? Below the text, we're given a histogram of the last digit of the student test scores, and on the vertical axis, we have the frequency, and on the horizontal axis, we have the last digit 0 through 9. We're also given 4 possible choices as our answers. For choice A, the digits are uniformly distributed. There is no evidence of grading bias or irregularity. For choice B, the digits are not uniformly distributed. There is strong evidence of gradient bias. For choice C, the digits are uniformly distributed. However, there is strong evidence of grading bias, and for choice D, the units are not uniformly distributed and also there is no evidence of grading bias. So, for the first part of this problem, we need to identify the frequencies of each digit, and we can actually read these frequencies directly off of our histogram. So, looking at our histogram, if we look at the last digit of 0, we see that it has a frequency of 10. For one, it has a frequency of 6. For 2, it has a frequency of 5. For 3, it has a frequency of 7. For 4, it has a frequency of 5. For 5, it has a frequency of 1246, it has a frequency of 6. For 7, it has a frequency of 8. For 8, it has a frequency of 4, and for 9, it has a frequency of 7. So that's going to be the answer for the first part of our question. And again, that just comes from directly reading these values off of our histogram. Next, we need to test the claim that the last digit of the scores are uniformly distributed. And for this, we're going to use a chi square test. So, to start off, we're going to write down our hypotheses. So, for our null hypotheses, H knot. We say that all digits. Or Equally likely. So this is going to correspond to a uniform distribution. And for our our alternative hypothesis, HA. That's going to be that the digits. Are not. Equally likely So this is sort of the opposite of what we have for our no hypothesis. And so now we actually need to perform a chi squared test. So, the first thing we need to do is to determine our expected value for when all the digits are equally likely. And so, we're gonna call that expected value. E. And this is going to be equal to the total number of frequencies. And so, if we add up all the frequencies that we just named, they all add up to 70. So, we're going to have 70, and that's going to be divided by the total number of categories, which we have 10. So, the digit 0 through 9 is going to give us 10 categories. So we'll have 70 divided by 10 as our expected value, which this is going to be equal to 7. So, if our digits are uniformly distributed, we expect a frequency of 7 for all the digits. So now we're going to use a chi squared test to compare observed frequencies to our expected frequency, to see if the difference is likely due to random chance. Now recall your formula 4 chi squared. And that's going to be equal to the sum. Of the quantity of O minus E in quantity squared divided by E. Where O here is the observed frequency and E here is our expected frequency. And so we'll need to calculate chi squared for each of our observed frequencies, and then add them together, and this is best done in a table. And so We have gone ahead and placed this table on the screen, and it consists of several columns. In the first column, we have the digits, which are 0 through 9. Next we have the observed frequencies, which we label as O. And just recall that in sending order going from 0 to 9, we have the frequencies of 10, 6575, 12, 684, and 7. And for our expected frequency E, in this column, we have the value of 7 for all of our digits. And then in the next column, we have the quantity of O minus E. And starting with the digit of 0 and going to the digit of 9, we have for that column 3 minus 120, -2, 5, -1, 1, -3, and 0. For the next column, we have the quantity of O minus C and quantity squared. And so again, starting with the digit of 0 and going to 9, those values are 9140425, 11, 9 and 0. And then finally, in the last column, we have the quantity of O minus E in quantity squared divided by E, and again, going from digit 0 to 9 in order. The values are 1.2857, 0.1429. 0.57140. 0.5714, 3.5714, 0.1429, 0.1429 again 1.2857 and 0. So now to calculate a value of chi square, we just need to add up all the values in that final column. And when we do, we get a value of 7.7143. So, we now need to use this chi squared and test it against the critical value that corresponds to a 5% level of significance. So to do this, we first need to determine the number of degrees of freedom or our data set. So we'll label that as DF and recall that the degrees of freedom is going to be equal to the number of categories, which in this case is 10, then we'll have to subtract 1 from that. So the degrees of freedom is going to be equal to 9. So, using a chi square distribution table, With a significance level of 5%, and a degree of freedom of 9, we see that the value of chi squared for this case is going to be 16.919. And so now we need to compare our critical value here of chi squared to the one that we actually calculated. And we see that our calculated value is actually less than our um critical value, so 7.7143 is less than 16.919. Which means that we fail to reject the null hypothesis. Therefore, there is no significant evidence to suggest that the last digit of student test scores are not equally likely. Therefore, the data does not indicate any grading bias or irregularity in how the scores are assigned. So, based on this information, we see that the correct answer choice for this problem is answer choice A. The digits are uniformly distributed, there is no evidence of grading bias or irregularity. So I hope that this explanation has been useful, and I'll see you in the next video.

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Key Concepts

Frequency Distribution

Chi-Square Goodness of Fit Test

Random Sampling

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