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Considerpaint-drying situatiorNhich dnyingescsdecimenonmally distributedThe hypotheses Ho:73 and Ka: M < 73 arebetested Lusingmanmom Sample of nodsenationsHon ma...

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Considerpaint-drying situatiorNhich dnyingescsdecimenonmally distributedThe hypotheses Ho:73 and Ka: M < 73 arebetested Lusingmanmom Sample of nodsenationsHon many Standaro devia tions (of X) below the standard deviations72.37 (Round your answedecimalaces )(b) If % = 72.3, what the conclusion using 0.004? Calculate the test statistic and determine the P-value (Round vour test statistic two decimal olaces anop-valedecimal places )DValueState the concluslon the problem context;Reject the null h

Consider paint-drying situatior Nhich dnying escsdecime nonmally distributed The hypotheses Ho: 73 and Ka: M < 73 are betested Lusing manmom Sample of n odsenations Hon many Standaro devia tions (of X) below the standard deviations 72.37 (Round your answe decima laces ) (b) If % = 72.3, what the conclusion using 0.004? Calculate the test statistic and determine the P-value (Round vour test statistic two decimal olaces ano p-vale decimal places ) DValue State the concluslon the problem context; Reject the null hypothesis There not sudicient evidence conclude that the mea drying time than 73_ Do not reject the null hypothesis There sufficient vidence conclude that the mean drying time less than 73_ Do not reject the null hypothesls There not sufficlent evidence conclude that the mean drying time ess than 73. Reject the null hypothesis There sutiicient evidence conclude that the mean drying time less than 73. For the test procedure with 8(70) 004, what B(70)? (Round You ansiver decimal places: ) (d) If the test procedure with sdecimens Msed ~hat neceslamc ensute that B(70) 017 (Round You 4nsiver the next whole number:) (e) If a 0.01 test Used 100, what the probability of = type etror Mhen 767 (Round Your ansher decima Dlaces



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Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? (e) Interpret your conclusion in the context of the application. Use the expected values $E$ to the hundredths place. Psychology: Myers-Briggs The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and $\mathrm{F}$ refers to feeling. $$ \begin{array}{l|c|c|c} \hline & \multicolumn{2}{c} {\text { Personality Preference Type }} & \\ \cline { 2 - 3 } \text { Occupation } & \multicolumn{1}{c} {\text { T }} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\ \hline \text { M.D. } & 77 & 82 & 159 \\ \hline \text { Lawyer } & 118 & 94 & 212 \\ \hline \text { Column Total } & 252 & 267 & 519 \\ \hline \end{array} $$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the $0.01$ level of significance.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

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.

Following is a solution to number 25 comparing to means for the amount of time lost due to hot temperatures Compared to the mean number need an amount of time lost due to disputes from superiors attitudes in the workforce. And the first part of this, all we're gonna do is just verify that the mean and the standard deviations for the two datasets are in fact 4.86 and 3.18 for the temperatures And then 65 and 288 for the attitudes. And it says use a calculator, so I'm using the T I T four and if you go to stat and then edit, you can see already put those numbers in. So L one represents, I think that's the temperature column and then L two is the attitude problem. And if you go back to stat and then cowpoke and then one bar stats, one variable stats the list, I'm just doing L one first and that is where we get that 4.86. And the standard deviation where this s is that's up 3.18. So that's where I get these numbers here, and then we'll do the same thing, cal one of our stats, but this time we're gonna do second to for L. Two, and we calculate that, and that's where we get the mean as 6.5, this X bar here, and then the standard deviation, we're not looking at signal, we're looking at essence is a sample sample standard deviation about 2.88 So that's where we get this 2.88 So we have verified that those are in fact the numbers, and now we're gonna do the two sample t test with the significance level of point oh five. Before we do that, we need to figure out what the um alternative hypothesis would be, because we already have the data. We can just punch that in after that and it's going to be a two tail not equal to test because it says one way or the other is just as are the two means different and whenever it doesn't specify which one is greater, you just assume that it's a two tail meaning not equal to so not equal to is going to be our alternative. Okay. And then there are no I didn't write it down but the Noles that they're equal to. Okay. So if we go to stat and then go over to tests and it's the two sample T test and since we already have the data, I'm just going to use the data instead of the summary stats because it's more accurate that way list one will be L one list too will be L to the frequencies can just stay as one and then the alternative the new one, let's change that to not equal to the pool is going to be no unless it's otherwise stated. And then we're going to calculate and this gives us everything. So the t. You know if you want you can put that in there. But really all I care about is this P value the P value is about 0.317 So I'm gonna write that down so the p value is equal to 2.317 And what we do is we explicitly compare the P value with our alpha value and this time the p value 0.317 is greater than our alpha value. Which means we fail to reject the null hypothesis. So any time the peabody is greater than alpha you fail to reject. If it's less than alpha than you actually reject. H not so if we fail to reject this null hypothesis, that means there's not enough evidence to say that these two means are different, or in other words, these two means appear to be the same, so I'm gonna write that out, or I'm gonna type it out because it's a little quicker. That way, I'm gonna, right, there is not sufficient evidence to suggest that the meantime lost due to hot temperatures is different from the meantime lost due to disputes from superiors attitudes in the workforce. Okay, so that is our two sample T test for these two population means.

We want to conduct a pair differences test at the output equals 1% confidence level testing and claim that sample or rather population means that our A. Is greater than expire. Be given the data for A and B. Below here assuming the mound shaped in symmetric distribution. As you can see, I've already computed D. Bar the mean the difference is 12.6 and equals five. And a standard deviation differences 22.66 on the right. Using the appropriate formulas, we proceeded the five steps below to compute this PDT first we check the requirements, evaluate hypotheses because the distribution shape it's appropriate to the students distribution degree of freedom is n minus one equals four. Null hypothesis is not equal zero. Alternative nuclear than zero and alpha equals 00.1 significance. Next our test statistic is T equals D. Bar over SD over route and or 1.24 from the tea table. This gives p value between 0.25 point 125 Thus we can conclude That thing is greater than α.. So we fail to additional hypothesis, meaning we lack evidence, moody is greater than zero


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