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Q1. The following data consists of the length of service calla in minutes and the number of components repaired Units 109 119 149 145 154 166 Minutesvi) Find 62 ,...

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Q1. The following data consists of the length of service calla in minutes and the number of components repaired Units 109 119 149 145 154 166 Minutesvi) Find 62 , 04, Var(e), Var(8), Var(62), Var(j4) and Var(P4); where / is expected service time for call which needed for components_ vii) Test the hypotheses_ a = 0,20, 30, 40; Ho: B = 0,5.0, 10.0, 15.0; Ho: &? = 0; Ho: Y4 = 60 and Ho: 60 at S% and 2.5% level of significance Also find the confidence interval ineach case.viii) Also estimate

Q1. The following data consists of the length of service calla in minutes and the number of components repaired Units 109 119 149 145 154 166 Minutes vi) Find 62 , 04, Var(e), Var(8), Var(62), Var(j4) and Var(P4); where / is expected service time for call which needed for components_ vii) Test the hypotheses_ a = 0,20, 30, 40; Ho: B = 0,5.0, 10.0, 15.0; Ho: &? = 0; Ho: Y4 = 60 and Ho: 60 at S% and 2.5% level of significance Also find the confidence interval in each case. viii) Also estimate the simple parabola model Minutes Bi. Unites Bz. (Unites)2 + ix) Calculate multiple correlation coefficient (R?) for viii) and comment with iv). x) Find 02 , 04, Var(fo), Var(81), Var(82), Var(62), Var(04) and Var(P) for viii); where HIS expected service time for call which needed for components Xi) Test the hypotheses, Ho: /o ` 0,20,30, 40; Ho: Pt 0,5.0, 10.0, 15.0; Ho: 0" 0: Ho: Y4 = 60 and Ho: 444 60 at S% and 2.5% level of significance Also find the confidence interval in csI each case; with ~amscanner



Answers

Focus Problem: Meteorology The Focus Problem at the beginning of this chapter asks you to use a sign test with a $5 \%$ level of significance to test the claim that the overall temperature distribution of Madison, Wisconsin, is different (either way) from that of Juneau, Alaska. The monthly average data (in ${ }^{\circ} \mathrm{F}$ ) are as follows. $$ \begin{aligned} &\begin{array}{l|llllll} \hline \text { Month } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } & \text { May } & \text { June } \\ \hline \text { Madison } & 17.5 & 21.1 & 31.5 & 46.1 & 57.0 & 67.0 \\ \hline \text { Juneau } & 22.2 & 27.3 & 31.9 & 38.4 & 46.4 & 52.8 \\ \hline \end{array}\\ &\text { Juriedu }\\ &\begin{array}{l|cccccc} \hline \text { Month } & \text { July } & \text { Aug. } & \text { Sept. } & \text { Oct. } & \text { Nov. } & \text { Dec. } \\ \hline \text { Madison } & 71.3 & 69.8 & 60.7 & 51.0 & 35.7 & 22.8 \\ \hline \text { Juneau } & 55.5 & 54.1 & 49.0 & 41.5 & 32.0 & 26.9 \\ \hline \end{array}\\ &\text { What is your conclusion? } \end{aligned} $$

Part one. The estimate on row is minus point 238 and the standard error is point 113 The T statistic is minus 2.12 with a P value of 0.38 So we have a we have evidence for serial correlation and the test is significant at the 5% level. In part two, I will not show the regression results for the whole equation. I will report just the value of beta too, which is the coefficient on the lock of riel gas price. All our estimate is 13 Sorry. Minus 13 0.96 and P W method gives beta to hat of minus 15. Applying 04 The P W estimate of beta two is more statistically significant. It is significant at the 5% level compared to you. The 10% of the old L s estimate. Thanks. In part three, we will obtain the new we West standard. Iran's using legs of 14 and eight. Remember that changing the legs does not change the estimate off. Beta beta too are the same for all types. They are 13 point knife five with one leg. The standard error Newey West style is six point 641 and the T start is minus 2.1 with four. Lex, the standard error is 7.167 and the T value is minus one point 95 and lastly, with eight legs. The Senate error is 7.881 and the T statistic is minus 1.77 Actually, I have the P value, So the P value for the first T statistic is let's write it here. Um, 0.4 for the second standard error estimate is 0.0 six. And for the last one. Okay, we have 0.8 So, as you can see, when the T statistic decreases in absolute value, as the legs increase, the estimate becomes less significant. We see that the P value increases. Let me write that down. In conclusion, at the leg of one, Beta two is significant at the 5% level, but using legs of four and eight mixed beta two, significant at the 10% level

We have the following matched pairs and we want to conduct a sign tested matched pairs, testing the claim P does not equal five at alpha equals 50.1 significance. This question is testing an understanding of how to implement statistical non parametric texts. In particular the scientists and matched pairs. We proceed the steps A through D to solve so first and a. We see our alpha and hypotheses alpha is 0.1 H and r S p 0.5 AJ is the claim P does not equal five and B. We compute the test at so your population A to B. We have the following signs thus and equals 20. The total number of pluses and minuses and X is the number of plus over. The total number 10/20. Thus our Zenon is x minus point 5/2 0.0.25 over and a zero because 00.5 minutes 0.50 Thus from a normal distribution, r p value is one because the probability at Z is greater than 0.5 times two is one. Thus, we conclude for this test that we fail to reject because P is greater than alpha, which means that we lack evidence to support the alternative hypothesis.

91 Which problem? 20 Which in a, um, you nothing is equal to. And one times me one plus one times Mewtwo, which is 118.6 Sigma Squared, which is 298.28 and the Central Division the squared off this value, which is 17.2, seven or eight for coaching be, um for A is equal to 1.5 and be equal to 2.75 So we will make the same calculations here instead. One, we will use 1.5 and here is 2.75 So, um is equal to 20 to 8 to 91.25 and the standard the variance is 1898.53 and the standard deviation squared off. This value is equal to 43.57 to 1 using the same concept but using different uh, so we use A is equal to 50 and B is equal to 1.5. So mu l is 15 plus 1.5 times you won, which is 92.45 This is a linear function that in your function is a plus. plus the X. Okay, so this is the mean and the variance is 1.5 square seen my one square, which is 151.29 And the standard deviation is the square root off this value which is 20.3.

Now, what do we have in this question? They're saying let X be a random variable that represents that every daily temperatures in the month of July in a small town in Colorado. Okay, now the ex distribution has a mean off 75 F. Okay, So the mean, the mean is 75 F. Okay. And the standard deviation sigma is approximately 8 F. Okay, Now, a study was conducted over a span of 20 years. That is 620 July days, and we are given a table off those increase from that study. Okay, So what are the columns that we have? Let's just look at the columns this we're also going to use in order to find the chi square statistic. Okay, so here it has, given that new minus three sigma is less than equal to X, which is less than noon minus two Sigma mil minus two sigma. Okay, similarly, here. I think that was me. Minus two Sigma minus sigma. Then there is mu. There is mu plus sigma, and there must be mu plus two sigma. Okay, these are Sigma's, this Sigma Sigma Sigma. Okay. Less than equal to less than equal to less than equal to less than equal to less than equal. Okay, then these are the low elements. And now we will write the upper limits. This is going to be Sigma. This is going to be immune. This is going to be mu plus sigma. This is going to be mu plus two Sigma, and this is going to be mu plus three sigma. Okay, so what is happening over here is we are having arrange the frequency. The frequency count for the number of days when the temperature waas between I mean minus three Sigma and me minus two Sigma. Right? When that when we can say that. Okay, if I see this normal distribution. Okay, let me just draw this normal distribution here. This is the mean. Okay, this is one standard deviation away. This is two standard deviations away. And this if I extend the graph, this is going to be three standard deviations of it. Right? So all of these categories are basically the frequencies off. How many values lie? Let's say between the first categories. View minus three. Sigmund U minus two Sigma so mu minus three. Sigma is 123 That is this point to me. Minus two signals this point. What is the frequency that lies between this? The number of values that live in this region, then in a moral values that, like between U minus two sigma and you mind a sigma, right? That is this region. So in this way, this is the study off all the different regions. All right, so this is the first column. The second column is actually the values. What a Sigma Sigma is eight. Right? So what is happening over here is now. They have just given us the values. When excess between 51 2. 59 between 1559. Then when excess between 59 to 67. Then when excess between 67 to 75. Then when excess between 75 to 83. Then when access between 83 to 91 Then when excess between 91 to 99. Okay, so these are the temperatures, right? What does the rains when X minus three sigma We do. We get 51 degrees, and when we do X minus two sigma, we get nine degrees. So what was the frequency or what was the number of days where the temperature waas between these two values? Right. So this is just a study off the number of values that are between two and three. Standard deviations away to the left, then one and two standard deviations away to the left and so on. Okay, now we are. I think we are also given the expected percent from normal have. Okay, now, what is happening is they're expecting this to be perfectly normal, right? This distribution to be perfectly normal. So this is the third column. Okay, So what should be the values? If it isn't be perfectly normal. It should be 2.35%. 2.35% for the first category. For the second one, it should be 13.5%. Okay, then it should be 34%. Then again, 34%. Then it should be 13.5% again. And then it should be 2.35%. Because the normal distribution is symmetry. We can see that the values here are symmetric, right? 2.35 13.5 34. Then in the decreasing order, 34 13.5 and 2.35 Okay, now this is expected. Okay, but what are the observed values? What are the observed observed values now? The observed value that there were 16 days when the temperature was between two and three. Standard deviations away to the left. Then for the second category, it was 78 days. Then it was 212 days. Then it was 221 days, then was 81 days. Then it was 12 days. Okay, now this addition is given to us a 6. 20 between that. This is our sample size in this. Is our sample size in Okay? No. What is going to be the expected value? In order to find the Chi Square statistic, we also need to find the expected values expected values E. Now, what is the formula for? This was the part A. We understood what the top of the first three columns are saying. Now, how do we find the that is the expected values this is given by the sample size. The sample size, which happens to be in in our case, which is 6 20 multiplied by the probability off each category. The probability or the proportion, right? The probability off each category. Okay, so let's say if this world this distribution work, uh, to follow a normal distribution, then this would be are expected probabilities in column three. Okay. Okay. So let's find the expected values for the first category. There will be now, use my calculator. Okay, So the expected value for the first categories. 2.35% of 6. 20. So point zero 235 the two into 6. 20. This is 14.7 41 57 14.57 Then it is starting 0.5% of 6. 20 0.135 in 26 20. There is 83.7, 83.7. Then we have 34% of 6. 26. 20 in two 0.34 This is 210.8 210.8. Then this is again 31. 34% of this shoot. Again. We 210.8. That this is 13.5% which is 83.7 again. 83.7. And then this is 2.35% which is 14.5 14.57 Okay, these are expected values. Now, in order to find the chi square statistic I need to find for all the cells, I need to find the value off. Oh, minus C. That is the observed value minus the expected value. Whole square upon the expected value. And then I need to sum them all up so that I get the guys were statistic for my entire problem. Okay, so this will be the column for individual for individual chi square values. Okay. All right. So let us just do what we saw in the formula. The difference between observed and expected. So this is 16 minus 14.57 We square this square 1.43 and divide this by 14.57 So this is 0.14 zero point went four, then the difference between 83.7 and 78. So this is 5.7 square and divided by 83.7. The 6.3 18 So I can write. This is 180.39 Then the difference between 212 and 210.8, we square this 1.2 square and divide this by 210.8. So this is 0.68 But I can write this at zero point 007 Okay, then the difference between 21 210.8. We square this and divide this by 210 point eight. This is 0.493 organizes 0.5 0.5 0.5. Now the difference between 83.7 minus 81. We square this and this is divided by 83.7. This is 0.87 0.87 And then there is a difference between 14.57 minus 12. We square this 2.57 and divide by 14.57 So this is 0.4533 0.4533 Now I have the individual Christ question districts. Now all I have to do is add them all up. So this is 0.14 plus 0.39 plus 0.7 plus 0.5 plus 0.87 plus 0.4533 So this is 1.5603 This is 1.5603 All right, so my high square value over here is 1.560 three. All right? Now, in order to find the P value, what I need to do is find the degrees of freedom. Degrees of freedom. DF is given by the formula. Number of categories, number off categories minus one. Okay, how many categories do we have? You see here the first If you look over here, we have 1234566 categories, right? So what is going to be our answer for degrees of freedom? It is going to be six minus one or I write this as Fife. Now, you can either use a chi square table to find the value, or you can use a chi square calculator or any statistical software, so I'm using it online. To hear 1.56 is my price square statistic, which is one point 5603 1.56 zero. Create on The abuse of freedom is five now, what is my Alfa when Alfa in this case is 0.1 Right. My Alfa is 0.1 as the developed significant is 1%. So this is 0.1 And when I calculate this, I find that my P value is 0.90 My p value is 0.90 Now I can see that my P value is greater than Alfa. Hence I say that I will fail to reject mine. L hypothesis. I fail to reject my null hypothesis. H not Okay now what was the hypothesis? What was my hypothesis here? Mine l hypothesis Waas. Okay, the I think we forgot to write the null hypothesis. My little hypothesis would be that the distributions are the same, right that the normal distribution O r Let's just say the average daily july temperature follows a normal distribution panel Hypothesis would be that the daily July temperature temperature follows a normal distribution. Okay, this waas my null hypothesis. What would be my alternative hypothesis that the normal distribution doesn't fate The daily July temperatures. All right, So what exactly I am I going to say I say that I failed to reject my null hypothesis, meaning that I I do not have enough statistical evidence. Enough statistical evidence to suggest that the daily July, the daily July temperatures and the normal distribution Okay, uh, do not have enough sufficient enough statistical evidence to suggest that the normal distribution that the normal distribution Let me just write it like this. Uh, just a moment. Okay. That the normal distribution that the normal distribution doesn't fit doesn't faked. Yes, the daily July temperature distribution. Okay, distribution. This line is going to be a answer. So I will say that I do not have enough statistical evidence to suggest that the normal distribution does not fit or doesn't fit the daily July temperature distribution. And this is how we go about doing this question.


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