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61 IRead problem 8.28 in your textbook Fill in the missing word HAn obrervation is considered to be influential 26+4) = 20+42_o303 13 No observations havela Hlevera...

Question

61 IRead problem 8.28 in your textbook Fill in the missing word HAn obrervation is considered to be influential 26+4) = 20+42_o303 13 No observations havela Hleverage point greater than 0.308. Thus, there do not appe:r to be any mpoxenyationsRead problem 8 29 in your textbook FFilllin the missing wordAn observation i8 influential if the Cook distance falla ab2ve the J0" percentile ofthe F Ertrbuton Iitunli*+1=4 +W5,47m-0+439050+7385 F85 T50e 0 87725 Obeeruation #8 hal 4 Cool , Ditance 0f 2

61 IRead problem 8.28 in your textbook Fill in the missing word HAn obrervation is considered to be influential 26+4) = 20+42_o303 13 No observations havela Hleverage point greater than 0.308. Thus, there do not appe:r to be any mpoxenyations Read problem 8 29 in your textbook FFilllin the missing word An observation i8 influential if the Cook distance falla ab2ve the J0" percentile ofthe F Ertrbuton Iitunli*+1=4 +W5,47m-0+439050+7385 F85 T50e 0 87725 Obeeruation #8 hal 4 Cool , Ditance 0f 2440433 Which % greater Ihan 0 8772 Ehs Ppeertaation ES 2



Answers

Refer to Problem 13. The following distribution lists the first digit of the surface area (in square miles) of 335 rivers. Is there evidence at the $\alpha=0.05$ level of significance to support the belief that the distribution follows Benford's Law? $$\begin{array}{lrrrrrrrrr}\text { First digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \text { Frequency } & 104 & 55 & 36 & 38 & 24 & 29 & 18 & 14 & 17 \\\hline\end{array}$$

In this problem, we'll see how taking out that 11.88 from exercise, one point for the one that never fell into any of those intervals, how that affects our calculations. So when we take out that high number are mean is going to come down and it's going to go down to 4.195 And the standard deviation is also going to be lower at 1.442. Now we'll construct some new intervals, one standard deviation above the mean and below the means. So four 195 minus 1.442 and 4.195 plus 1.442 will give us a range of values that go from 2.753 Up to 5.637. And remembering empirical rules, says 68% of our data will fall in here. If it's normally distributed, let's count up how many of our 39 data points fall within these two values. And it appears that 25 out of the 39 values fall in that interval, which equates to around 64%, which is close to that empirical rule prediction. Now our next one, we're going to go two standard deviations above and below the means. So we're going to multiply our standard deviation by two and add and subtract that result onto our mean. Mhm. Which will give us a range of values that go from 1.311 up to 7.079. The empirical rule suggests 95% of our data should fall with that interval, let's see how well our data fits that prediction. 36 out of the 39 data points fall within this interval and that equates to 92.3%, which is pretty close to that 95% prediction. And lastly, let's go three standard deviations above and below the mean. And this will give us a range of values that go from negative .13, 1 up 8.5-1. And now all 39 data points fall within three standard deviations of the mean, which is very close to what the empirical rule predicts of 99.7%.

We want to conduct a pair differences test at the output equals 1% confidence level testing and claim that sample or rather population means that our A. Is greater than expire. Be given the data for A and B. Below here assuming the mound shaped in symmetric distribution. As you can see, I've already computed D. Bar the mean the difference is 12.6 and equals five. And a standard deviation differences 22.66 on the right. Using the appropriate formulas, we proceeded the five steps below to compute this PDT first we check the requirements, evaluate hypotheses because the distribution shape it's appropriate to the students distribution degree of freedom is n minus one equals four. Null hypothesis is not equal zero. Alternative nuclear than zero and alpha equals 00.1 significance. Next our test statistic is T equals D. Bar over SD over route and or 1.24 from the tea table. This gives p value between 0.25 point 125 Thus we can conclude That thing is greater than α.. So we fail to additional hypothesis, meaning we lack evidence, moody is greater than zero

So we're giving a problem that we have to find at the 5% significance level. The sample size is told to be of 10. So and is equal to 10. Which implies that are Significant value or critical value is at five and 10 and we're losing one till test. So we have our critical value of W. C. is equal to 44. Now when we take a look at this test that we're supposed to do it's one tailed. However which tail is it? We are asked to say that one is more effective than the other. So mm everything is mostly positive here and we find what are our values? Mhm. Use the following data set which gives the additional sleep in hours obtained by 10 patients who used love level high school. So I mean high bromide. So that's gonna be so the control group minus the other one or not. The control group. So what we'll just do here is calculate the both ends since I'm not sure if it's a To a left tail or right tail. So we'll go ahead and find out that right tail is at 44 and then the other tail is going to be 10 times 11 divided by two. So that's gonna be 55 minutes 44 Which is gonna be 11. Very handy. Okay. So if it's less than 11 won't reject and if it's greater than 44 will also reject. Sound good? So um we got 10 students at the positive values so 1.9 plus 0.8 plus 1.1 plus 0.1 was 4.4 plus 5.5 plus 1.6 plus 4.6 plus 3.4. And we get 23.4 for our w nut. Okay. And or critical value that just leads us straight in the middle. So at alpha equals .05 We failed to reject the null hypothesis and this was just a right tail test that are evaluation is insignificant. So now if we tried it again at the 1% significance level, mhm at point a one that changes our values to be mhm uh 50. So we got 50 right here and then we do 10 tons 11 divided by two. So 55 -50, which is going to be five. And once again 23.4 still lands us right in there. So we failed to reject the my hypothesis once more.


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