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FIOm the three imuges fepresenting three tounds of >lection and repopulation YouI dala fTOt Ihe count ofthe Sunvors Make surc Jour tolal sardsprt found adds up 2...

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FIOm the three imuges fepresenting three tounds of >lection and repopulation YouI dala fTOt Ihe count ofthe Sunvors Make surc Jour tolal sardsprt found adds up 25 or you have miscountedSeedRound SurvivorsRound Survivor}Round Survivors106.3 Lo, 4033} " S 15 10.667Kidney BanBlack BcanFnto BcaD(ireen FeaWhat trend did You sce sunvivors the above dataAccording to the data on the table; trend for survivorship is like both white bean and green pea have better survival rate than the other three

FIOm the three imuges fepresenting three tounds of >lection and repopulation YouI dala fTOt Ihe count ofthe Sunvors Make surc Jour tolal sardsprt found adds up 25 or you have miscounted Seed Round Survivors Round Survivor} Round Survivors 106.3 Lo, 4033} " S 15 10.667 Kidney Ban Black Bcan Fnto BcaD (ireen Fea What trend did You sce sunvivors the above data According to the data on the table; trend for survivorship is like both white bean and green pea have better survival rate than the other three colors Wbat the null hypothesis tor experiment _ Null hypothesis -if all five types of beans and peas are grown in & controlled way the average survival of white bean (10.3) and green pea (10.667) are much higher For 4 dof at p-0.05 value 9.488 which is much lower than the calculated chi-square value 771 approx mated hence we should accept out null hypothesis Use the null hypothesis calculale the expected number of seeds that should survive out of 25 that are left at the end of each round Write these into the table below . Now . perfonn Chi-Square Test using Round 3 $ sunivors data (since represents the most dramatic outcome alter thrce generalions of selection) Observed- (Obs-Exp) X2-[Qbs_Exnk Expected (-0.333) 0.111 0.011 (-0.667) 0.445 0.667 (-0.333) 0.111 0.333 Seed Observed Expected White Ban 10.333 0.667 0.333 Black Hcan Pinto Bean Orcn /c1 10.667 (4.3331 18.775 1.760 Add up the final X? from each TOw t0 get the EX? What are the Degrees of Freedom (DoF )for your experiment DofF (M-1) (5-1) =4 (DofF number of variables outcomes (seed tPes) Whutc Kiduc}



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Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table $A$ - 3 with df equal to the smaller of $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

First one. There are 248 families do not want the apples at any price Or two. So distribution is not continuous, There is focal points and rounding for example. Many people report on powder and either 2/3 of a pound or one and a third pounds. This the fact that the distribution of quantity demanded is not continuous violates the underlying assumption of the topic model. Yeah. Which is the Layton error has normal distribution, but we will still explore the Tobin approach in this context. Yeah. It may work better than the linear model for estimating the expected demand function, do you? And Along with Part eight. The estimates from their topic and L. S.. Models are reported in the same table or the tablet model. The price variables ali them are Statistically significant at the one level. The sign over these prize coefficients are in accordance with the demand theory, the own price effect is negative and across price effect is positive. Cross price is the price of this substitute good, which is regular apples part Let's do part 6. 1st part six we will obtain their fitted values and we find that the ranged from 27 8 798 to you. 3.33 at five. The null hypothesis is later one plus beta two equals zero. This is something you can easily test regardless of your statistical package. Yeah. Yeah, you should get a small T statistic about minus point you and a P value of buying eight. So we are unable to reject the North part seven. The squirt correlation between it is E call B. S and it's fitted value is about .04 and that is there are square hard eight. Given the linear model estimates, even this result, we find that the old LS estimate are smaller than the top bit estimate. And in terms of our square oops, you compare the goodness of fit between the two models we look at. There are square mhm. There are squared of the topic. Model is still smaller, slightly smaller than the old LS model and serve. We can conclude that the topic is yeah, no better than the old LS. It doesnt suite the data better. The Last Part, Part nine. The statement is simply incorrect so you could run into a uh counter example. We have valid price effects, but we cannot explain much of the variation in the dependent variable. It's simply difficult to estimate the demand for a fictitious product.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct. And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified. And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there. I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two. That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you. A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there. But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value. That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A. Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis.


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