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Which at the statistical tabies shquid Z-distnbutiomUSA?t-distribution F-distribution LI pt(s)] Supmm Answer Tries 0{1What are deqrees freedom assaciated with the t...

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Which at the statistical tabies shquid Z-distnbutiomUSA?t-distribution F-distribution LI pt(s)] Supmm Answer Tries 0{1What are deqrees freedom assaciated with the test statistic? Numerator df; Denominator df: pt(s)] Submit Answcr Tries 0{1Calculate the test statistic, Submi Answer Tries 0{3Pt(s)]Whlch Intcrva In thc tablc contalns thc valuc for thc test? (Usc thc tablc on thc coursc sitc and conflrm vour answcr with softwarc:) value D.OD1 0,001 p-Val p-value L05 ,05 p-value P-value 0.25 p-vaiue

Which at the statistical tabies shquid Z-distnbutiom USA? t-distribution F-distribution LI pt(s)] Supmm Answer Tries 0{1 What are deqrees freedom assaciated with the test statistic? Numerator df; Denominator df: pt(s)] Submit Answcr Tries 0{1 Calculate the test statistic, Submi Answer Tries 0{3 Pt(s)] Whlch Intcrva In thc tablc contalns thc valuc for thc test? (Usc thc tablc on thc coursc sitc and conflrm vour answcr with softwarc:) value D.OD1 0,001 p-Val p-value L05 ,05 p-value P-value 0.25 p-vaiue 0.25 Pts Submt Answver Tries D{1



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The variance for each of Student's $t$ -distributions is equal to $df/(df -2 )$. Find the standard deviation for a Student's $t$ -distribution with each of the following degrees of freedom: a. 10 b. 20 c. 30 In summary: d. Explain how this verifies Property 5 of the $t$ -distributions listed on page 414.

In this question. I have been given a table. All right, Andi, I have to calculate. I think the mean is to be calculated by using that table. So if I use the table to find the mean the mean off those 32 these are 32 numbers, right? Yeah. These are ready to farmers focus. So their main turns out to be 14.83 4th expert is 14 0.834 My end is equal to 32. Have I been given the standard Division? Yes. Population standard Division 6.2 Sigma is equal to 6.2. And the claim? What is the claim? It takes smokers too. Quit permanently. Is there enough enough evidence to reject the claim that the meantime, it takes smokers to quit smoking permanently as 15 years? Okay, the mean is 15 years. So immune is equal to 15. This is going to be my null. And my alternative hypothesis will be that Mu is not equal to 15. All right, I need I need to know what I'm going to use. I'm going to use the sad statistic and I have all the information that I'm going to require in order to calculate this, so I'll simply put in the formula 14.83 minus 15 upon 6.2 by route off 32. Now I have to use the calculator to calculate this Just a moment. So this is going to be 14.83 minus 15 divided by 6.2, multiplied by route off 32. This is minus zero point. This is minus 0.155 This is my zed statistic. Okay, Now, if I calculate my people first of all, this is going to be a two tailed test because there's a not equal to sign. And what is my Alfa level? My al 50.5 So if this is a two tailed test night in these two tails, my total areas 0.5 which means in every single tail it will be 0.25 So let me do one thing. Let me find the critical value for 0.25 and this value turns out to be 1.96 which means that since it is symmetrical, this is minus 1.96 and this is plus 1.96 And since my Z value is between this interval, my Z value falls somewhere around here minus 0.155 I will fail to reject minor hypothesis. And if I want to find the p value for this, my P value for this is going to be P Valley for minus for the zero point. Uh, sorry for minus 0.1 point 55 The P value for this is just a moment in a zero point 88 My P value is turning out to be 0.8 ID, 0.88 right. This is my p value. I have used the people you calculate over you, and I can see that my p values greater than Alfa. So from here also, I can say that I will fail to the Jackman and hypothesis. So this was my null hypothesis. I will say that I do not have enough evidence to see or I do not have enough evidence to reject the claim that the meantime it takes the smokers to quit smoking permanently is 15 years. This is going to be my answer

Let us read this question. A company that next cola drinks states that the mean caffeine content per 12 ounce bottle of cola is 40 mg. So my mean is 40. This will be my it's not or my Nelly apotheosis. What will be my alternative hypothesis? It will be mu is not equal to 40. All right, You want to take this claim during your test, you find the random sample off 12. So my end is 12th. Sorry, my random sample of 2012 ounce bottles. So my end is 20. My end is 20 22 Alone's borders of cola has mean caffeine content of 39.2. So my expert is 39.2. Okay, it is 39.2 mg assumed that the population is normally distributed and population standard deviation Sigma is 7.5. So sigma is seven point fight 7.5 mg and Alfa is equal to 0.1 and Alfa is equal to 0.1 Can you reject the company's claim? Okay. What is the first thing we want to identify? The null and the alternative hypothesis. We have done that the be part says find the critical values and identify the rejection regions for standardized test statistic Z. Okay, what are going to be my critical values now? This is a two tailed tests and my al 50.1 So let me be clear that this is a two tailed tests and my al 50.1 So, in each one of these tales, how much area should be there? It should be 0.50 point 005 Okay. And if I find the critical values, they are going to be plus minus 2.575 plus minus 2.575 So this is plus 2.575 and this one is minus 2.575 And what is this Z statistic that I get if I put in the formula 39.2 39.2 minus 40 minus 40 upon sanity, vision is 7.5 upon a Route 10. What is N. N. Is 20. So this is route 20. This turns out to be minus. This turns out to be minus 0.5 minus 0.5 eight. Okay? And I can see that this value falls somewhere around here. So this is not in the rejection region. Excuse me. So I will fail to reject minor hypothesis now decide whether to reject or failed. Yeah. So we will fail to reject the null hypothesis. And if you want to interpret the decision, what was on al hypothesis? It was that mean is 40. So we will say that we do not have enough statistical evidence to say, uh that the mean caffeine content 12 ounce bottle off cola is different or anything other than 40 mg, and this would be my answer.

Now. This time I have a figure in the textbook and my critical zed value is 1.285 So it is So it is something like this, right? And this is 1.2 it for at 1.285 It's my first one is minus one point 31301 So my part is minus 1.3. So according to the figure, my rejection region is this. This is my rejection region. And since this 1.3 falls somewhere around here, I will fail to reject minor hypothesis. Okay, The second one is 1.203 now 1.203 will fall just to the left of this. So again in part B also, when it is 1.203 I will again fail to reject minor lie apotheosis in part C. It is 1.280 in part C. It is 1.280 This will be one point 280 On this, I will see that it will again fall to the left. It will be very close, but it will fall to the left. So again I will fail to reject and 1.286 in case of 1.286 It will fall in the rejection regions somewhere around here. So I will in part D when I have 1.286 I will reject my edge. Not Andi. He's would be my answers.

So we believe that the mean number of ounces is 12 ounces and alternately, it doesn't say anything about trying to find if they're under filling or over filling. So we're going to use the not equal to and do a two tailed test. Now they found that the mean was 12.19 ounce. And we had a standard deviation of 0.11 ounces. And we have a sample size of 36. And so again, we're going to assume that 12 is the center and we're getting 12.19 And since it's a two tailed tests are also going to be finding both of these together will be our P value. So, well now I'm going to do this finding the P value and not doing it with the finding the critical T value. So what's the likelihood of sampling If the mean is actually 12 and getting 12.19 for a mean? Now we want both ends of the tail, so we're going to double that. So we need to convert that to a T. Value and we would have 35 degrees of freedom and we'll take our 12.19 minus 12 and then divided by the standard deviation or the square root of an and when we do that calculation that numerator becomes 0.19 divided by and then 0.11 and that looks like that's divided by six. And that comes out to a test statistic holy moly that test statistic comes out to be uh make sure I didn't type anything in in incorrectly. We get a test statistic of 10.36 and if we find that p value that's going to be approximately zero. So we actually have strong evidence what evidence against the null since this is way less than 5% which is our significance level. And so we would say that the mean the main seems to be different is different. Mhm. From 12 ounces now and again our p value is very very close to zero. It's a little bit bigger than zero. Now, it asked, does this does this end up show evidence in the way they were to the question that does it say that the people are being cheated and no they're not being cheated. Not cheated. Uh It looks like they're actually getting more than 12 ounces. So we didn't do that significance test for more than but it doesn't appear as though they're getting cheated. What?


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