5

Point) Matt thinks that he has special relationship with the number 5. In particular; Matt thinks that he would roll a 5 with a fair 6-sided die more oiten than you...

Question

Point) Matt thinks that he has special relationship with the number 5. In particular; Matt thinks that he would roll a 5 with a fair 6-sided die more oiten than youd expect by chance alone Suppose pis the true proportion of the time Matt will roll a 5_(a) State the null and alternative hypotheses for testing Matt' $ claim. (Type the symbol "p" for the population proportion_ whichever symbols you need of = not = and express any values as fraction e: P = 1/3) Ho = p = 1/6Hap > 1/

point) Matt thinks that he has special relationship with the number 5. In particular; Matt thinks that he would roll a 5 with a fair 6-sided die more oiten than youd expect by chance alone Suppose pis the true proportion of the time Matt will roll a 5_ (a) State the null and alternative hypotheses for testing Matt' $ claim. (Type the symbol "p" for the population proportion_ whichever symbols you need of = not = and express any values as fraction e: P = 1/3) Ho = p = 1/6 Ha p > 1/6 Now suppose Matt makes n = 30 rolls; and P-value comes up times OUI 0f the 30 rolls. Determine the P-value 0f the test: (c) Answer the question: Does this sample provide evidence at the percent level that Matt rOlls (Type: Yes or No) more often than youd expect?



Answers

Apply the chi-square goodness of fit test, using either the critical-value approach or the P-value approach, to perform the required hypothesis test. Loaded Die? A gambler thinks a die may be loaded, that is, that the six numbers are not equally likely. To test his suspicion, he rolled the die 150 times and obtained the data shown in the following table. $$\begin{array}{|l|cccccc|} \hline \text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Frequency } & 23 & 26 & 23 & 21 & 31 & 26 \\ \hline \end{array}$$ Do the data provide sufficient evidence to conclude that the die is loaded? Perform the hypothesis test at the 0.05 level of significance.

Okay. In this question, you want to use the scientists to find What their data indicating that 50% of adults 10 to not trust their government is accurate. Um, in this test and development, selected 2000 engine too. Mm hmm. American adults And found that 1000 200 and 40 three did not trust, you know. Okay. Yeah. So this is a right to yeah. Yeah. Prices. And our process. Yeah. Is that the deed five? Can I get your access? It's not All right. It's like she quit. So acting 0.05. Yeah. Of the of significance. All right. We can find that the Critical value using normal areas table is 1.64. Mhm. Since this is in large someone. Yeah. And that's really that's right. You can use From one check out statistic because you know what? He came closer than five. Mhm. And number two. Okay. It has. Thank you. What should get less. Well, then they created. Yeah. That's what I've finest. Mhm. Okay. Yeah, Yeah, Yeah. Sperry two. Sure. Okay. She'll give us a lesson six. Mhm. Since scare them. Sears five. We cannot reject my advances because there is insufficient evidence That the Germans significance that 50 of adults do not trust the level

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.

Okay, so in this question, were given that X is equal to 6062 on physical too. Okay, 5938. Plus, that's 6062 which will equal to 12,000. We're else given. P is your five on Alfa are those significance is your is your five all right? Okay, so now we can calculate our proportion, which is X over end, which is 6060 to over 12 passion, which is equal to 0.50 52 Yeah, it's now the hypothesis is given by it's not to know that, but says P is your five that alternative hypothesis H A P is less than 2.5. Okay, Now can calculate our Z test statistic, which is question might be cover squared. Q p off on a kid, thank you is equal 21 nice people. P is your for five. Yeah, So if you plug all these values and we'll get a Z value of 1.14 and to now you can get our critical values. So z half a is equal to negative 1.645 from the table that is our C critical value Andi R P value, which is just the p probability that g less than 1.14 which is just equal, does your point is 7 to 9 on. Since this is greater than our critical value, which is 0.5 you can fail to reject our no hypothesis h not.

What do we have this time? Okay, so there is a gambler who thinks who has his suspicions, that the dice that is given to him the die is actually unfair. So that I was rolled 300 times and it's frequencies were recorded. And that data is given to us. Let's look at the table. What is the data that we have? We have the outcomes. Okay. What are the possible outcomes off a die? One, 23 four, five and six. There are six possible outcomes. Okay. And what are the observed frequencies? The observed frequencies. The observed frequencies are 60 to 45 63 62 45 63 30 to 47 51 30. Do. 47 51. All right. Now, what exactly is the question? They're asking whether the data indicates that the die is unbalanced and we have to use a one person level of significance. Okay, One person level of significance. Which means that our Alfa is 0.1 What will be a null hypothesis? Arnold? Hypothesis will be that that die is balanced, the die is balanced. And what will be the alternative hypothesis? The alternative hypothesis will be that the die is unbalanced. The die is unbalanced. All right, now, how do I know if I die is balanced or not? Well, if a die is balanced, every outcome, every outcome will be equally likely. Every outcome will be equally likely. So let's say that if the rural probability is one and there are six possible outcomes, what is the probability for every outcome, it will be won by six, which is 0.167 All right, so my ex probability Let me just write a column for probabilities, for probabilities. What will be the probability? The probability is will be 0.167 for all of them. These are the expected probabilities. This should be the distribution that it should follow. If it is balanced 0.1 67 0.167 All right. No. I'm going to conduct a chi square test. And what is the first step to conducting a chi square test? Calculate the expected values. Calculate the expected values for all categories for all yeah categories. All right, this means what? How do I calculate the expected values? Expected values will be for every category. We're going to perform this calculation. It will be sample size sample size multiplied by the probability multiplied by the probability off each category off each. Yeah, category. Okay, so if we go over here, this is the column for the expected values. The expected values. All right. What is the total sample size? It is 300. So what will be the expected values? They will be 302.167 or 50 right? Since they are all equally likely. Ideally, I should have 50 as the frequency for all of them. 50 50 50. All right, So I have the expected values. Now, what is the next step? Now I want to calculate the Chi Square statistic. How do I do that? For every category I'm going to perform the calculation observed value minus the expected value. I'm going to square it, and I'm going to divide this by the expected value. In the end, I'm going to some of the entire column. And in this way I get the Chi Square statistic. Let us put this formula into action. This is going to be chi square values. These will be the case for values for all the categories and I'll send them all up in the end. So this is going to be 60 to minus 50. I will square this 12 square is 1 44 I divided by 50 which is 2.882 point 88 Similarly, over here it'll be 25 by 50. So this will be 0.5 then I have 13. Square is 1 69 1 69 divided by 50 t is 3.38 Then I have 18 squared, divided by 50. This is 6.48 Then I have nine divided by 15, which happens to be 0.18 Then I have one divided by 50 which is 0.0 to 0.2 Now I will add all of these up. So this is 2.88 plus 0.5 plus 3.38 plus 6.48 plus 0.18 plus 0.2 And this will give me my chi square value as 13.44 Now that I have my chi sequestered escape in order to find the P value. What I need is the degrees of freedom. And how exactly do I find the degrees of freedom? It is given by the formula? Number of categories, number of categories that I have in my problem, minus one. How many categories do I have? If I look at the table, I have six different outcomes. So there are six categories in this question. So I will find that my degrees of freedom is six minus one. Or I can say that this is equal toe fight. Now that I have my chi square value and my degrees of freedom, I confined the P value. So I'm using online tool over here My significant level 0101 And I find that my P value is 0.195 My fee value is 0.195 What was my Alfa? My Alfa was 0.1 I can see that my people who is greater than Alfa hence I fail to reject I I feel to reject h not that is my null hypothesis. So in the end, what is going to be my conclusion? I say that I do not have enough evidence in just a moment. Just a moment I think we are doing. We're making some sort of a mistake because this die is clearly imbalanced, right? My chi square statistic is 13.44 That makes sense. My degrees of freedom was five. My degrees of freedom is five. So this is saying that the result is not significant. What I'm getting over here is that the result is not significant. The result is not significant if the result is not significant, which means I will fail to reject my age, not which will mean just a moment. I think we're making a little bit of a mistake. This is clearly this is clearly unbalanced. So if we check our calculation here, expected values are 50. This is 2.88 This is right. This'll is right. This also look straight. These air also rates are addition turns out to be our addition turns out to be 13 13.44 are degrees of freedom is five and I'm getting my P values 0.195 which is greater than Alfa which means that this is not significant. I failed to reject my age. Not so. What was my age? Not over here my age not said that the diet is balanced and the alternative hypothesis says that the die is unbalanced. Okay, although this to my eyes this looks clearly unbalanced. But at 0.1 at Alfa is equal to 0.1 I say that I do not have enough evidence. I do not have enough evidence to suggest that that die is unbalanced. Although this clearly looks unbalanced to me. But at 0.1 level of significance, I cannot say that this is unbalanced. If it were all physical to 0.5 I would say that he has. This is definitely unbalanced. But at 0101 this is not unbalanced because P is greater than Alfa. All right, so this is how we go about doing this question.


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