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Business Administration student wants determine there 3 significant relationship between corporates revenues (in milliions) and actual profits (iIn billions} She su...

Question

Business Administration student wants determine there 3 significant relationship between corporates revenues (in milliions) and actual profits (iIn billions} She surveyed tne top 10 Fortune 500 companies: Test the following data at the .05 level. What Is the value of the test statistic? Round the thousandths placeKa AeeanMcet nn Untdhealut 75 HclaTou KotorRevenue224216 205166Pralits 13.613.0

Business Administration student wants determine there 3 significant relationship between corporates revenues (in milliions) and actual profits (iIn billions} She surveyed tne top 10 Fortune 500 companies: Test the following data at the .05 level. What Is the value of the test statistic? Round the thousandths place Ka Aeean Mcet nn Untdhealut 75 Hcla Tou Kotor Revenue 224 216 205 166 Pralits 13.6 13.0



Answers

Consider again the relationship between the sales and profits of Fortune 500 companies that you analyzed in Exercise 40.
a) Find a 95$\%$ confidence interval for the slope of the regression line. Interpret your interval in context.
b) Last year the drug manufacturer Ell Lilly, Inc., reported gross sales of $\$ 9$ billion (that's $\$ 9,000$ million). Create a 95$\%$ prediction interval for the company's profits, and interpret your interval in context.

For this problem were given the regression analysis output, or at least some of it for he sales and profits of a randomly selected bunch of Fortune 500 companies and for part A we arrest are asked to test a hypothesis to see if we think there is a significant association between sales and profits. So I know hypothesis is that the slope is equal to zero, which is to say that there is no association between sales and profits. And the alternative hypothesis would be that the slope is non zero, and therefore there is an association between sales and profits. And so to conduct our test of our head pa theses, we want the T value and the P value so we can easily calculate the T value from the information we have. It's equal to the slope minus zero, divided by the standard error of the slope, and this comes out to 12 point 33 And if you look up a P value for a T score of 12.33 with 77 degrees of freedom ah, you get something that's approximately zero. You get a very small P value and therefore based on this very small P value, we would reject the no hypothesis and conclude there is evidence of an association between sales and profits. And then, for a part B were asked if we think that the company's sales are useful predictor of its profits and were asked to use the R squared value and the standard deviation of the residuals. So based on the R squared value of 66.2% that means that about 66.2% of the variability in profits can be explained by our model on sales, which means that there are other factors in in the profits and the standard deviation of the residuals is 466. So that's 466 ah 1,000,000. So even for a company that's making a $1,000,000,000 worth of sales in a year, ah, you'd only you don't expect to have based on this slope, you'd only expect to have about $100 million of profits, which means that the standard deviation on the residuals is very big compared to profits that some of the large companies might be making, which suggests to us that this is not a particularly useful predictor for the profits. Sales is maybe a moderately good predictor for the profits, but not ah, highly, um, not a highly accurate predictor of the profits. And maybe we shouldn't even see good. There may be just a moderate.

We want to perform a hypothesis test and we have a sample. This is from the I'll call it finance instead of one. But we have the 110 people from the Finance uh, group and the mean for those with income was $48,537 and the standard deviation of that sample was $18,000 of quite a large standard deviation. And then from the business analytics group we have I'll call them just the business ends a B. We have a sample size of 30 so significantly smaller and the standard Excuse me, the mean of that group was $55,317 and we have the standard deviation of that group being $10,000. And we want to do a hypothesis test, and so are null hypothesis, as we want to find out if this group of finance is significantly lower in salary than the business analytics. And so we would assume that the mean for the finance is equal to the mean for the business analytics. And alternately, we would think that the mean is less than the other mean, and we're going to use a 5% significance level. I believe that's what it said to use. And we need to now perform our test so we will be assuming that the difference between these two is zero. So we're going to assume that the mean for finance minus the mean for the business analytics is equal to zero. And we can see that from our sample that we're actually going to get the difference between the X bars being being negative. So it's going to be down here someplace, but we want to find out if it's significantly different. So we next have to find our test statistic, and our test statistic is going to be a T value now. Are degrees of freedom that we're going to use is going to end up being the We're going to use a conservative estimate. You know, there's that big formula that we can use, and we can use software to help find a more accurate one. But if we take the lesser of the two sample sizes and take one less than that, that is a conservative estimate for what the degrees of freedom is. It's not that we can use that awesome formula or software to help us figure out what the actual number of degrees of freedom is. But we're going to take the difference between the two lips 537 minus 55,317 and then we're assuming that the difference is zero. So the difference we got minus the difference we're assuming over. And now we have that big standard deviation that we have to deal with. And we will take the, uh, let me move that to there so I can see our standard deviation is 18,000. For the first group, we square it, find its variance and divided by its sample size, which I believe was 1 10. Yep. 1 10. Yeah. And then the standard deviation of the other I can see is 10,000 dollars squared and then divided by it. And I believe that sample sizes 30. I just peek. Yep, that's it. So now I have not calculated this, but the first thing I'm going to do is find out what this difference is just like, kind of market on this graph. So 48,537 minus 55,317. That difference is negative. Negative. $6780. So that's what the X bar for the finance minus the X bar for the B is. And we are going to find that area, and that will be our P value. Yeah. Okay, I know. We know, test statistic wise, that we're dealing with that being a T value of zero, we're going to figure out what that test statistic is in that t distribution. All right, so we have to take that number and divided by the square root of 18,000 squared and see if I put it in up zero. Yes, I did. Divided by the sample size of 1 10. Plus that 10,000 squared, divided by the sample size of 30. And that gives me a test statistic. And that's one of the I think the next question is of negative 2.705 and it goes eight. And so that's what that test statistic is right there. That t value is negative. 2.7058 and now we want to find the P value. So that was part B and part C. we want to find What is the likelihood of getting that test statistic like that or smaller in that distribution? And so my I'm going to use my T c D f on my calculator. So I'm plugging in my negative, like 1000 for the lower limit. The upper limit is this value right here, which is that negative 2.7058 and our degrees of freedom again. Our conservative estimate is 29. Yeah, and that gives me a P value of 0.56 So it's a very small P value. So now what's our conclusion? Since this P value is less than 5% and it was quite a bit less than than 5% we have strong evidence to reject the null mhm no, and conclude that the mean salary for the finance is less than mhm, the mean for the business analytics. Since the P value is significantly smaller okay than 5%. And remember, what you're really looking at is if the main difference is actually zero, we do expect variability. The likelihood of getting this kind of difference of $6780 is extremely unlikely to happen if the main difference is really zero. So it only happens less than 1% of the time if the true difference is actually zero and and we are all done.

In this problem. We are looking at how much business at a garage comes from former patrons and we believe it is 68%. But we are going to test if it is actually less than 68%. As is detailed in part a. And we're going to use the 1% significance level very similar to the last problem. We just did. So 0.1 and 0.99. R. Z score is going to be negative two point uh 3-6. So the .1 I'm going to put right here negative 2.3 to 6. N. E. Z score less than that is going to be room to reject are null. So then we have to calculate our Z score. Our sample is 114 out of 200. Which comes out to be 57%. And we want to calculate our Z. score and see how different is that. We can see it's 11% points. Which is pretty significant. But our sample size is small. So it could just be due to random chance. So let's see what happens when we calculate our Z score. So we calculate our Z score, We get negative 3.335. Which is going to be all the way down here again. All right. So it is in the rejection region. So, we will reject the null. And our reason why is because negative 3.335 is less than negative. 2.3-6. Okay, So it is less than R. Z score for our rejection region. Now, part B. It wants us to calculate the Um the actual p. value. So we're gonna use our figure 12.2 to calculate that. So I'm I'm gonna go down to negative 3.3 Which is 0.0004. That is easily less than one. So there we reject the null as well. So if you're using the observed significance or the P. Value you're going to compare it to you're going to compare it to your significance level. If you're using the Z. Score you're going to compare it to the Z. Score for the rejection region in order to prove that.

So this is a pretty quick problem. The first thing you would want to do is create a dummy variable for Roz Neck that you really have to do, just depending on the language programming language you're using the computer application you're using. Sorry. And, um, now we just want to explain what our beta three coefficient meetings. Beta three hat means so Beta three is the same thing as the coefficient of Ross neg. And what this means is, if the return of stocks is negative between 1988 to 1990 then what is that relationship between this and, um, he salary of our CEO? So what we have is this is equal to negative point two three, about negative 0.23 So notice that we have a log salary and this log basically as a proxy, this eyes an approximation of a percent change. So what we're saying is that when we have a negative return on stock, all right, if the return on stock was negative between 1988 and 1990 then the CEO salary dropped by about 22.6%. Um, for whatever the given levels of sales and Ara, we are on DDE. We know that this is ah, significant because of our P value of 0.4 and that's less than 0.5 So at the 5% significance level, um, we have a ah, we have a signal. We have a significant coefficient value. So against summarize between 19 88 in 1990 if the return of stock was less than zero than CEO salary, um, was predicted to be if I could spell predicted as 22.6% lower, Um, then if it was positive, were equal to zero, um, for the given levels of sales and return on you, whatever that stands for.


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