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You suspect that your east-facing portion of your Macintosh apple orchard is producing less apples than your west [ facing portion_ You decide to conduct_ study of ...

Question

You suspect that your east-facing portion of your Macintosh apple orchard is producing less apples than your west [ facing portion_ You decide to conduct_ study of your apple trees to determine the average production of apples per tree as measured in bushels for both sections. Conduct hypothesis test at 5% level of significance to determine if the East facing section is producing significantly less apples per tree compared t0 the West facing section. Use the table below:Sample size Average Stand

You suspect that your east-facing portion of your Macintosh apple orchard is producing less apples than your west [ facing portion_ You decide to conduct_ study of your apple trees to determine the average production of apples per tree as measured in bushels for both sections. Conduct hypothesis test at 5% level of significance to determine if the East facing section is producing significantly less apples per tree compared t0 the West facing section. Use the table below: Sample size Average Standard of trees amount of deviation] Bushels West 55 28 East 50 What is the p-value? 0.05<p<0.1 p-0.0021 P=0.0042 0.025<p<0.05 p-O. 9979



Answers

An appropriate null hypothesis to test whether the
trees in the forest are randomly distributed is
(a) H0:M
25, where M
the mean number of trees in each quadrant.
(b) H0:p
0.25, where p
the proportion of all trees in the forest that are in Quadrant 1.
(c) H0:n1
n2
n3
n4
25, where ni is the number of trees from the sample in Quadrant i.
(d) H0:p1
p2
p3
p4
0.25, where pi is the actual proportion of trees in the forest that are in Quadrant i.
(e) H0:pppp ˆˆˆˆ 1 2
3 4 0. , 25 where pˆ i is the proportion of trees in the sample that are in Quadrant i.

Now, this is question number nine from the textbook. Over here. We have a diagram. Okay, We have a t I displayed on the left hand side, and we're going to use this display to answer all the questions we have to make. Remember that we are using the 0.5 significance level. Now. What is going to be Arnell? An alternative hypothesis. We are using the data to test the claim that they form a population having a mean less than 4.0 MVPs. Mean is less than four point zero MVPs M. B. P s. This is what we're testing. So this becomes our alternative hypothesis, right? So what is going to be our null hypothesis? Arnel hypothesis ends up becoming mu greater than equal to 4.0 MVPs, 4.0 MVPs. Okay, what is Artie statistic from the diagram, we can see that Artie statistic is minus zero point minus 0.366 And what is the P value? The P value that is given to us in the diagram is 0.3579 0.3579 Okay. What was the significance level? What was Alfa, it was 0.5 Okay, this was our Alfa for analysis. Okay, So using this Alfa, we see that our P value is much greater than Alfa. So we fail to reject our null hypothesis. This is our null hypothesis. So we fail to reject it, right? So we say that we fail to reject Anil hypothesis and that there is not sufficient evidence to support the claim that the Sprint Airport data speeds are from a population having a mean less than four MBPs. Right? So we failed to reject h not we fail to reject h not and again in words. We will say that there is not sufficient evidence to support the claim that the Sprint Airport data speeds are from population having a mean less than 4.0 MVPs and these would be my answers.

So in this question, we are given a board off a T I 84 close. So if we look at the board, were given the value off RZ statistic were given our P value which yes, 0.443 were given the mean off the first sample population X one, the second sample population X two and the sample size and so on. And what we're asked is to use the display to make a decision to either. So this is the now hypothesis. So we're asked to make the decision to either reject or fail to reject the now hypothesis at the level off significance given here. We're asked to make the decision using the standardized test statistic as well as the P value. So if you look here are people us 0.3 which is less than 0.5. So using our p value, we can reject the now hypothesis. Now, rZ statistic is 2.956 Approximately so. So for our Z statistic, we know the critical value for Z for a two tailed hypothesis test at Alfa Aiko's point or five is 1.96 And so since our value of C is more in our critical value of C. Again, we can reject no hypothesis. This is the answer to the question.

Problem. 17 8. Each note is that new one is smaller than or equal commute. Each one is equal to me. One is bigger than you, so the degree of freedom is the minimum off. Anyone minus one into my storage is 18 and 14, so it's equal toe food correspondent to critical value with offer. April 2.451 tail so T is equal to 1.761 So the actual reason contain old values. Great around 0.761 So the test statistic is equal to x one bar minus X to bar. So for 97 year zero minus 4 to 000 over square. Note on 8800 square over 19 plus 51 Use your square over 15, which is ableto 3.19 point. So it's a very all the tests statistic in the rectory does not have pointy inject. Three point is smaller and it's bigger than 1.761 So we reject then the hypothesis

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.


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