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10_ The mean yield of corn in the United States is about 120 bushels per acre. surey of 40 farmers this year gives Sanple mean of 124.3 bushels per acre_ We want to...

Question

10_ The mean yield of corn in the United States is about 120 bushels per acre. surey of 40 farmers this year gives Sanple mean of 124.3 bushels per acre_ We want to know whether this good evidence that the national mean this year is not 120 bushels per acre_ Assume that the fatTers surveyed are an SRS fTOm the population of all cormercial cOrn growers aIdthat the standard deviation of the yield in this population is 10 bushels per acre. Is there enough evidence from the sample to suggest that t

10_ The mean yield of corn in the United States is about 120 bushels per acre. surey of 40 farmers this year gives Sanple mean of 124.3 bushels per acre_ We want to know whether this good evidence that the national mean this year is not 120 bushels per acre_ Assume that the fatTers surveyed are an SRS fTOm the population of all cormercial cOrn growers aIdthat the standard deviation of the yield in this population is 10 bushels per acre. Is there enough evidence from the sample to suggest that the population mean is not [20 bushels per acre?



Answers

The mean yield per acre of soybeans on farms in the United States in 2003 was 33.5 bushels, according to data obtained from the U.S. Department of Agriculture. A farmer in Iowa claimed the yield was higher than the reported mean. He randomly sampled 35 acres on his farm and determined the mean yield to be 37.1 bushels, with a standard deviation of 2.5 bushels. He computed the $P$ -value to be less than 0.0001 and concluded that the U.S. Department of Agriculture was wrong. Why should his conclusions be looked on with skepticism?

All right, I'm gonna show you how I put the data in here. I just saved the same spreadsheet as I had for the previous, um, previous question. Now we want to investigate the relationship between acres planted and tons of sweet corn. So I'm gonna change this two acres, and I'm gonna change this two tons. And I noticed that there are 123456789 10 11 12. There are 13 rose that I need, but I only have 123456789 10. Early of 10. So I'm gonna add a few more rose. Do insert. Rosier. And so, um 50 353 69 365 um, 21 144 34 187. 20 100. Juanito, 7483. Get all this gotta in here. 86. Okay, there's all of the data. Now. I had this information in here to calculate the standard air, but I don't think we need to do it in this question. Although I haven't really read the whole question yet, so I have a scatter plot year now. now it's No, it's not points on All Star Game. I'm just going to write, scatter plot of the data. There we go. Okay, so I'm supposed to include a scatter plot Linear coal correlation coefficient. That's the our value. Um, but most spreadsheets give you the R squared value. I noticed that the our value would be positive. Those who is going to be positive square root of 9.55 equals square root of 9.955 That's going to be a correlation coefficient of 0.977 Just strong upward. Uh, correlation. The line of best fit is written right here. It's 6.15 times x times the number of acres plus 13.3 and a statement about their meaningfulness. So again, the our value indicates a positive relationship, and it is strong, strong, ah, strong relationship. The equation is a prediction of Ah, if you knew how many acres it would predict how maney tons, um, could be reaped. So let's go on to part B. The question and be is how many ah puns can you expect to produce per acre? Well, I notice that the Y intercept is not zero. You'd really think that the Y intercept should be zero. Because if you plant zero acres, you're going to get nothing. Sell. I know in Microsoft Excel you can actually force the X enters the Y intercept to be zero. I don't know if there's a way to do that in here. Um, air bars format the data point. I don't see a way to do it in Google seats, but in Excel, it's like forced to be zero like a check. Uh, forced the y intercept to be zero um, linear exponential. If it just said proportional here, then that would work too. So that appears to not be a function in here. So I guess the best thing that weaken dio a word my linear ago, Where is my linear? There it is. Is ah, 6.15 But I would say that this line should actually be at the zero down here. Um, but the point is, about 6.15 would be the advice per acre planet. However, it does appear that for whatever reason, the smaller farms are somehow more productive than what you would expect. And I might discuss that with the sweet corn grower. If he wanted to hear it. Seems a little unusual. All right. Thank you for watching.

There are two bids to this question. The first one is, Is there a difference in the variability off county yields as measured by the standard deviation of the rate? I have been given two columns in the question. All I have to do is enter them with many types software. As you can see that I have done here, I have this posted the screen short over here. What is my null hypothesis? It is that the ratio of the sigma's is one, and the alternative is that the ratio of the symbols is not one. I am assuming that you know these are taken from a normal distribution. These samples are taken from a normal distribution, and hence I can use the F statistic. So over here you can see that I have applied the statistic and my P value is zero point 941 Hence, I can say that I cannot reject the null hypothesis, which means that these Sigma's are almost equal. According to this test, let's go to the next one the second bit. And this question says, is the main yield rate in Wisconsin significantly higher than in Minnesota. Well, here I am going to use a T test at test two. Sample t test Again. I have the same columns. What is going to be my random variable? Here it is. Mu w minus Mu m. This is my null hypothesis. This is my null hypothesis. And if I'm using my alternative hypothesis going to be that this minus, this is greater than zero. All right, I can see that my p value over here is 0.35 which means that I have enough evidence at Alfa is equal to zero point 05 my p values less than Alfa. Hence, I can say that I can reject the null hypothesis and I can say that immune W or the everything Wisconsin is statistically significantly higher than u M or Immune in Minnesota. This is how we do it.

We want to test the hypothesis New equal 694 against the alternative. New greater than 694 for n equals 40 sample mean expert equal 731 with standard deviation of the population 212 known and the test conducted at alpha equals 2120.5 confidence. This question is testing our knowledge of how to conduct a hypothesis test for population. Mean you where the standard deviation of the population segment is known to do so we're gonna rely on normal curves. First we calculate the test statistic, this is N equals x minus mu over sigma, whatever it. And you're not as our 694 in the null hypothesis and the Xenon is 1.10 plugging in the above numbers. Next for the confidence level, given we have two computer critical value for a right tail test. This is the critical Z score on a normal curve for which the area to the right is 20.5 Thus, from a Z table, we get Z C equals 1.6 point five. Thus, because 1.10 is less than 1.645 and it's not in the critical region, were able to conclude in part C, that we fail to reject a chart because then it is not in the critical region.

In this problem, we are asked to carry out a significance test um for the yield of read hard red winter wheat and were given that are null hypothesis is that it's about 44.8 Bushels Per acre? That's what it's been in the past. We would expect it to be the same. But we are trying a new method to see that if it will increase that. So that would be a one tailed test in which it is greater than 44.8. We want to make sure that we have a A random sample of land plots. Were testing that on um and it does tell us there's 35 independent plots. So we're going to assume those are randomly selected And our central limit theorem, we are greater than or equal to 30. So our our sampling distribution will be approximately normal. Um and we're testing at the 5% significance level. All right, So now um we have to calculate our Z score the first way we're going to test. This is using the critical value. We observed a sample mean of 45.4, We would expect it to Bend 44.8. We're going to divide it by our standard deviation of our sampling distribution, which is what we have right there. And we can calculate that Z score which is really the key part that we need. This should be 35 down here Divided by the Square of 35. And that gives us a Z score of two 0.2 one. It actually rounds up to two. It's A 2.22. Now we have to consider what our um Z score is going to be if we take um So what is the Z score if the area of our rejection region is .5 And everything to the left of that is .95 and we get that are Z score for doing area? The left would actually be Um .95 is going to be Um 1.645. So that's the Z score that we're comparing it to an hour, given Z score that we calculated is going to fall right in here. So it does fall in that rejection region. So we can say that since 2.22 Is greater than our 1.645. We reject the null. Now, part B, we're going to calculate the p value and we can do so using normal CDF. All right. RZ score was 2.22, we're calculating to the right, so all the way to infinity since it is a one tailed test And are mean and standard deviation for our standard normal curve is zero and 1 and we can do that with a normal CDF calculator And we get a p value of .0132, and then part C r p is less than our significance level 0.5. Therefore we will again reject the null. So it totally matches every time we've done this, it is matched up If our Z score is bigger than the Z score for our given significance level, we reject if R. P. value the area is less than the area that we have, so the area to the right of our value is .132. We just needed it to be somewhere less than .05 to fall in that rejection region on the standard normal curve.


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