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Suppose VOl roll Luo dice [00 times. Each time YOu record their diflerence (alwavs suhtracting the smaller one from the bigger one get positive clifference) The pos...

Question

Suppose VOl roll Luo dice [00 times. Each time YOu record their diflerence (alwavs suhtracting the smaller one from the bigger one get positive clifference) The possible values YOn get are 0.1,2,3,4 aud You record the frequency of each value in the following table:Difference of two dice Observed frequencyLet Your HUll hypothesis he that the dice an fair , Audlthc alternative hypothesis he that they are not fair. Using coniclence leve] of 0.1, test the null hpothesis by goodness-of-fit test.Hiut

Suppose VOl roll Luo dice [00 times. Each time YOu record their diflerence (alwavs suhtracting the smaller one from the bigger one get positive clifference) The possible values YOn get are 0.1,2,3,4 aud You record the frequency of each value in the following table: Difference of two dice Observed frequency Let Your HUll hypothesis he that the dice an fair , Audlthc alternative hypothesis he that they are not fair. Using coniclence leve] of 0.1, test the null hpothesis by goodness-of-fit test. Hiut : begin by completing table: I(x) 2 3



Answers

Someone claims to be rolling a pair of fair dice. To test his claim, you make
him roll the dice 360 times, and you count up the number of times each sum
appears. The results are shown below. (For your convenience, the chance of
throwing each sum with a pair of fair dice is shoun too.) Should you play
craps with individual? Or abserved frequencies too close to the
expected frequencies?

We are testing to see if a die is fair. So we roll it 60 times and record the results in this table and part A were asked how many times we would expect each face of the dye to turn up if it actually is fair. So if it is, the fare die. Each face has an equal opportunity and equal chance of turning up and their six faces. So the probability for each face to turn up is 1/6, and we multiply that times the number of times that we roll the die. And therefore we would expect each face to turn up 10 times out of 60 rules so we can make a column in the table, code expected. And then for each face of the die, he expected cancer. 10 and then in part B were asked if if we would use a goodness of fit homogeneity or independence test to test whether this is a fear die. So a fair die means that every face has a one in six chance of turning up. This is a theoretical distribution, and we're testing whether the counts in our experiment lead us to believe otherwise. So we're testing whether the counts are good fit for the theoretical distribution. Therefore, we're using a goodness of fit Chi Square test and for part C, we're asked to state our hypotheses. So the no hypothesis is that it is a fair die. Another way of saying that is that all die faces have an equal chance or more specifically, a probability of 1/6 turning up. And so the alternative hypothesis is that it is not a fair die. We can simply state that not all faces have a probability of 1/6 of turning up. And for D were as to check the conditions. So one condition is counted data and we have the counts for each face of the dye turning up so in our table. So this is condition is met. Another condition is independence and random ization. And so the counts for each face are independent of each other. When we were all the die repeatedly, each role is not affected by the other rules. So the condition of independence is met with respect your random ization. The no hypothesis is that it is a fair die. Fared. I is a theoretical concept and therefore any single roll of a die is intrinsically random. So we have met this condition and we also want to meet the expected cell frequency. And that is that we want any of these expected sale counts to be greater than five, and they're all attend. So this condition is also met. And for party wrist, how many degrees of freedom? So an is the number of categories three degrees of freedom is equal to n minus one. So n is the number of different categories, so each face of the die is a category, so we have six minus one, therefore five degrees of freedom. And then for F were as to calculate the chi squared statistic and the P value. So you, if you were doing this manually, you can. You can do it by expanding this table so you might have You might make another column that is observed, minus expected, and then the next column being observed, minus, expected squared until you have a chi squared statistic column, which is going to be like that. So observe minus expected squared, divided by expected. And let's just do face one together. So face one is observed is 11 expected is 10 and that comes out to one of her 10. So we would put 1/10 in the cell and then the Skype, the Chi squared statistic. It's simply the sum of this column, so it would be six. Number is the 1st 1 is 1/10 and this comes out to 5.6 now. I have done this in many tab, so this valley was calculated for me. I didn't go all the way through manually, and it also gave me the P value, which was zero point 347 and then a G were asked to state our conclusion. So with a P value of 0.347 that is a very large P value, and therefore we are going to fail to reject the no hypothesis, and you could summarize by saying there is not sufficient evidence to suggest that the die is not fair.

Little this is from 13.31. We're going to be using the chi squared goodness of fit test. We'll be talking about dice one day in particular. The no hypothesis is that to die is not voted and the alternative is that die he's loaded are alpha value is 0.05 and now we need to figure out the probabilities. So the probability of landing a die being ruled in landing on a one is the same, is the probability that he lands on to That. It lands on three for five and six When the Dye is not loaded which is 16 since we have a six sided die, The values are 1, 2, 3, 4, five and six and uh 150 trials were done and out of those this is observed And it was 150 trials 23 Times of London on one 26 on two 23 on three 21 on 431 on 5 26 on six. No, the expected counts is equal to the sample size times of probability, Which in this case it's 150 trials multiplied by the 16 for which is probably That die lands on one and simplifying we'll get 25, you would have to do the same thing for everything else. So This would be 25 as well and so on. So that's right that down. Okay now we need to find what chi squared is. So chi squared will be good to some value. We need to figure out where the degrees of freedom are and we need to find the p value which is denoted as probability that okay squared is greater than a certain value. It's equal to something. Okay. In order to get the chi squared value though, we need to figure out what will be actually using these numbers right here. So we need to input the observed values and the expected values in a website called stacked ology dot org. And after that, after we get the chi square value and we have the degrees of freedom or use the other website called Salt by single judge. Learning to get our p value. Okay, let's get to it. As you can see on this first part we have All the observed counts are inputted in the expected values and our chi squared test statistic is 0.88. As shown, We'll get your .8 or degrees of freedom is just the number of categories which are six And subtracted by one should get is five. And we need to find the probability that chi squared is greater than The Chi Square Statistic, which is your .8. We use the website called salt and as you can see we have in put it that we have a chi square distribution Degrees of freedom are five And the probably the X. is greater than or equal to 0.88 Is going to be 09 seven, 0.9. Now we need to interpret what is happening because pierre value Is equal to 097 which is growing there equals 0.05. We do not reject. Was it no, meaning the data do not provide sufficient evidence to conclude that die is loaded.

All right, We have a six sided die, and we're gonna roll it 120 times and were given some data about how many times we ended up with a one or two and so forth. So we have our outcome. We rolled the die 120 times and 15 of the 120 times. We ended up with a one 29 times. We ended up with a two 16 threes, 15 fours. We got 30 fives and we got 15. 6 is. And if we were to total up the frequency column, we will see the 120 rolls. Now. These are are observed frequencies we want to calculate are expected frequencies. Now, if you think about the fact that when you roll a die there's six sides. So the probability of getting a one is 1/6 and the probability of getting a to is 1/6 and the probability of getting a three is a 1/6 and the probability of getting four is one out of six. And the probability of getting five is one out of six, and the probability of getting a six is one out of six. So if it's one out of six would be comparable to what out of 120. So if I said one out of six is comparable to what Out of 120 you would find out that we would expect 20 ones and 20 twos and 20 of each of the outcomes. Now we're ready to run our hypothesis test. So now we're going to come back to this chart in a moment. But before we conduce that we have to write our hypotheses, So we need to construct a null hypothesis and an alternative hypothesis. And if we're expecting all of these to be equal, then we're saying the diess fair. So that would be our null hypothesis. And the alternative to that would be that the die is unfair. So in order to run this hypothesis test, we're going to have to use the chi square goodness of fit test, and that's going to require us to find a chi square test statistic. And to find that Chi square test statistic, we're going to sum up the column that we evaluate when we do observed minus expected squared over expected. So now we're gonna go back to our chart. So we need to do observed minus expected quantity squared, divided by expected. So that means we're taking 15 minus 20. Getting a result were squaring it and then we're dividing it by 20 Now. Faster, more efficient way is to get our graphing calculator out. Sorry about that. My calculator. Go. There we go. And I've placed my observed values. Enlist one. And I've placed my expected values enlist to now while I'm sitting on top of list three. I'm going to tell it to subtract the observed values. Enlist one minus the expected values enlist to I want to square that result and then divided by the expected values enlist to. And I'm going to get these values. So I'm gonna put my calculator away, and I'm going to copy those into the chart so you can get rid of how we were finding it. And we're gonna type in 1.25 4.5 0.8, 1.25 13 point no five sorry. And then 1.25 And to calculate our test statistic, we need to sum up thes values. And when we do that, our test statistic is going to be 13.6 now. I think it's time that we draw a picture of what's going on. So a chi squared distribution is going to be a distribution that is skewed to your right. We're going to record the chi squared values down the bottom and the probability density on the side. And we need to talk about our degrees of freedom. And our degrees of freedom is found by doing K minus one where categories minus one. Our categories are all the different outcomes. So we have 123456 different categories. So our degrees of freedom in this instance is five and the mean of a chi square. Distribution is equivalent to the degrees of freedom. So the mean of this distribution is going to be a five, and we always find our means slightly to the right of the peak. So we're gonna have the chi square value of five right here. And we just found our test statistic to be 13 6. So if you think of numbering are axis 123456789 10, 11, 12, 13, 13.6 is gonna be somewhere right around here and we're ready to find RPI value and R P value is going to be the area of the curve that's in the right tail beyond that 13.6. So to find RPI value, we're trying to find the probability that the chi square value is greater than that 13.6. So what will use is our chi square cumulative density function found in our graphing calculator. And the cumulative density function asks you for a lower boundary followed by an upper boundary, followed by your degrees of freedom. So when we type in our chi squares CDF, we're going to use 13 6 as our lower boundary, and our upper boundary is just going to be some number way high on the horizontal axis. So we're gonna say 10 to the 99th Power and our degrees of freedom we said was five. So again, we're gonna bring in our graphing calculator and to access that chi square function, we're gonna hit the second button and the distributions, and it's number eight on my calculator and I wanna be in my home screen. There we go, and I'm going to put my lower boundary 13.6 my upper boundary, 10 to the 99th Power and my degrees of freedom of five. So I'm getting my P value to be 01 eight 36 So what that's saying is, that is the area in that tail that's this area right here. 0.0 183 Now we can get this very same image and the same data. If we go back to our graphing calculator again, I'm gonna clear this out. And if you hit your stat button and you go over to tests, we conduce you a chi square goodness of fit test, which is letter D in mind and my degrees of freedom. I'd have to type in this five and notice you see the chi square value that we had our, um, if I scroll down the value we had for our test statistic and r p value show up right on our calculator and notice our picture. Very similar. All right, we've got this little area in the tail and our pique is close to that five. Like our picture. Okay, so now that we have done that, we're ready to make a decision. So our decision is based on whether your significance level is greater than your P value. Now we normally do a significance level of five. So we're gonna let Alfa equal five. But your most common ones are a five significance level or a 50.1 significance level. So if we run our test at a 0.5 significance level, we're asking whether 0.5 is greater than 0.183 And the answer is yes. And because it's yes, the decision is to reject the null hypothesis. So again, I'm gonna take it back to the null hypothesis here, and our decision was to reject that. So we're throwing that away so we can then conclude that the die is an unfair die. So our conclusion there is enough evidence to conclude based on observed roles, that this die is unfair. And that's the hypothesis test using the chi square goodness of fit

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.


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