The outcomes on a die are normally one to three, four, five or six. And in this particular problem, the author is drilling a hole in the dye and fills it with a lead weight and then rolls it 200 times and observes the frequencies. And the frequencies were 27 31 42 40 28 and 32. So those air referred to as are observed frequencies. So we want to run a hypothesis test at a 0.5 significance level to test a claim, and our claim is that the outcomes are not equally likely due to the lead weight being put into the die. So before we can run the hypothesis test, we're going to have to create a no hypothesis and an alternative hypothesis. Now you are no hypothesis are always that the outcomes fit the expected values. So we're going to say that are null hypothesis are that the outcomes are equally likely and our alternative hypothesis will be that the outcomes are not equally likely. So we're going to be running a goodness of fit test to see how well or if the actual observed roles fit what is expected And if they don't fit what is expected when you roll a normal die, then we're going to support this claim. So in order to run the goodness of fit test, we are going to have to calculate the Chi Square test statistic for this data. And in order to do so, we're going to apply the formula. The sum of observed minus, expected squared, divided by expected. So let's go back to our chart, see what we have and see what we need. So we have our observed frequencies. Now we know that we rolled 200 times, so if you add up the observed frequencies, you are going to get 200. Now, If this was a fair die, we would expect an equal number of ones twos, threes and so forth. So if we take that 200 we divide it by the six possible values, we will get 33 a third. So our expected frequencies for each number should be 33 1 3rd. So we're now ready to come up with our Chi Square test statistic. I'm going to extend out my chart, and this time I need to calculate the observed, minus the expected quantity squared, divided by expected. So I'm going to take the observed data for one. I'm going to subtract the expected data. The difference that I get. I'm going to square and then I'm going to divide it by the expected value or the expected frequency. And the fastest, most efficient way to handle this is to utilize a graphing calculator. So I'm going to bring in my graphing calculator, and I am going to clear out my list. Three before we start and I'm going to go to stat edit and, as you can see, enlist one. I've already placed the observed frequencies and enlist to I have placed the expected value off 33 a third. So for list three, we're going to sit up on top. We want to calculate calculator to generate these values, so we're going to tell the calculator to take each observed value from list one. Subtract each corresponding expected value from list too square that difference before dividing by the expected value from list, too, and we will get these decimals. So, for the sake of reporting it on our paper, I am just going to write down to three decimal places, so the answer for one should be 1.203 for two is 20.163 for three is point. There's 2.253 for 41.333 for 5853 for 6.53 And to calculate my Chi Square test statistic, I need to add all of the's values up. Well, I already have them in the calculator, and they're actually more accurate in the calculator because we've carried them out to more decimals. So if I could tell the calculator just to add up list three, I will have my Chi Square test statistic. So I'm going to quit out of my list, and I'm going to tell the calculator to sum up list three. And in doing so, I'm getting a Chi Square test statistic off 5.86 So that's one component off our chi square Goodness of fit hypothesis test. We still need to find some additional components, so let's now transition to finding RPI value. R P value is going to be the probability that Chi Square is greater than that test statistic. We just calculated. And to get a better handle of what that is, I recommend drawing a picture. So because we're running a chi square, goodness of fit test, we're going to draw a chi square graph, and the Chi Square graph is going to be skewed to the right, and the shape of the graph is dependent on the degrees of freedom, and the degrees of freedom are found by taking K minus one and K represents the number of categories that we divided our data into. And we have divided our data into 123456 different categories. So therefore, K is six and our degrees of freedom than is five. Now, the degrees of freedom also tells us what the average of our chi square distribution is. So our average is five, and you will always find your average slightly to the right of the peak of the curve. And this horizontal axis is called Chi Square. And what we're trying to find is the probability that Chi square is greater than 5.86 So 5.86 is going to be slightly to the right of that average, and we're trying to calculate the area of this curve to the right. Now, in order to calculate that area, it's most effective to use our chi square cumulative density function from our calculator. And in doing so, it's going to ask you for three pieces of information. The lower boundary of your shaded area, the upper boundary of your shaded area and your degrees of freedom. So for our problem, the lower boundary of the shaded area is our chi square test statistic. Now the upper boundary. Keep in mind that that shaded area continues on and on, so we're going to just pick a large number out in the tail, and I'm going to use 10 to the 99th, and our degrees of freedom was five. So I'm gonna bring in the graphing calculator again, and we're going to hit second. There's and number eight in this menu is the Chi Square cumulative density function. The lower boundary is 5.86 The upper boundary is that super high number 10 to the 99th Power. And again, our degrees of freedom was five. So we get a P value of 0.32008080 to 5. So what that's telling us is that the area in this curve makes up for approximately 32% of that chi square distribution. There's one more piece of information that we find useful in running these goodness of fit tests, and that is your chi square critical value. And the chi square critical value is found by looking in your chi square distribution table in the back of your textbook. And when you look in that table, you're going to find degrees of freedom down the left side of the chart and your level of significance across the top of the chart, and your level of significance is going to be characterized by the Greek letter Alfa. And for this problem, we are using a 0.5 level of significance to run our test. So in your chart across the top, you're looking for your level of signage significance of 0.5 and down the side. We're looking for our degrees of freedom of five, and when you find where those two rows and columns meet up, you will get your critical value of 11.71 So we have all the parts we need to complete our test. So let's just recap before we run our decision and our conclusion. Our test statistic is 5.86 Our P value is approximately 0.32 and are critical. Value of Chi Square is 11.71 Now It comes time to do the decision. When when we do the decision, there are two different methods of deciding the outcome of this test. We can either utilize the P value or we can use the chi square critical value. Either one. You do not have to do both. They will yield the same result if you elect to use the P value. What we're doing is we're comparing the level of significance to the P value. And if the level of significance is greater than the P value than your decision must be to reject the no hypothesis. So let's use the P value to determine our decision. So in this instance, the P value Sorry the Alfa was 0.5. The P value was 0.32 and we cannot say that Alfa is greater than the P value. Alfa ends up less than the P value. So our decision based on this data is to fail to reject the no hypothesis now if we wanted to use the chi square critical value and get that same decision. What I recommend is that we draw a chi square distribution and place your critical value on that curve. So are critical value was 11.71 And by placing that on there, you have broken the graph into two regions. The region in the tail is called your reject. The null hypothesis region and the other half or other part of the curve is your fail to reject the null hypothesis region, and you will then place your chi square test statistic on or into this picture. And our Chi Square test statistic was 5.86 and 5.86 would be back here because it's smaller than 11.5 So since our test statistic fell in the fail to reject region, our decision is fail to reject the no hypothesis. So again, you can do either method to come up with your decision one or the other. You do not need to do both, so we're now ready to make our draw Our conclusion there was insufficient evidence to reject the null hypothesis. Since we can't reject it, there is not in a evidence to support the claim. Let's go back to our no hypothesis and our alternative hypothesis. Remember, our claim was right here at the alternative. So since we can't reject this, we cannot support this. So there is not enough evidence to support that claim. So finally, there's one other question that we needed to answer here. And that question says, Does it appear that the loaded die is behaving differently than a fair die? And we would have to say that the loaded die does not appear to behave differently than a fair die, and that concludes our hypothesis test.