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For Ech statemcnt bclor, express thc null bypothesis Ho tnd thc altcmative hypothesis Hi in subulk_borm: Btqurt t0 net tnt corrtct rymbol (05 Py &} for tht Indi...

Question

For Ech statemcnt bclor, express thc null bypothesis Ho tnd thc altcmative hypothesis Hi in subulk_borm: Btqurt t0 net tnt corrtct rymbol (05 Py &} for tht Indicatrd paraiacter:() The mcan annual income of employecs who took statistics courec igTeatcr Ihan S60,000. #o (6) Tbe sandard deviation of human body lemperatures is cqudlto 0.62" Fy #o (c} Tbe proportion of homes with burglar alans js 0.56. #o (4) The standard deviation of duration times (in scconds) ofthe Old Faithful geyser is

For Ech statemcnt bclor, express thc null bypothesis Ho tnd thc altcmative hypothesis Hi in subulk_borm: Btqurt t0 net tnt corrtct rymbol (05 Py &} for tht Indicatrd paraiacter: () The mcan annual income of employecs who took statistics courec igTeatcr Ihan S60,000. #o (6) Tbe sandard deviation of human body lemperatures is cqudlto 0.62" Fy #o (c} Tbe proportion of homes with burglar alans js 0.56. #o (4) The standard deviation of duration times (in scconds) ofthe Old Faithful geyser is less Jhan 40 Ic= #o Use the information giten und the flowchart on Pg:& 0l the formula packet t0 find the P-valuc in eh of the following: (4) The test stalistic in # lefl-tailed test is 2-125. P-value (6)The test statistic in & right-Lailed test is 2 2S0 P-vulue For cach combination of F-value und decide whether you #ould reject Q1 fuil t0 reject the null hypathesi Wrte " reject H' QE "Tail to Tcject" Ior cuch (4)P-value (b)l Pvalue (elP-vulue 0S4 Td) P-valuc 0;u Tolinucting contidence H degreraofAreedumn ~disuibution; ! tlc samiple s1 i 44.Ihcn thc tumbct



Answers

(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value and identify the rejection region, $(c)$ find the test statistic $F,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal. If convenient, use technology. The table shows the salaries of a sample of individuals from six large metropolitan areas. At $\alpha=0.05,$ can you conclude that the mean salary is different in at least one of the areas? (Adapted from U.S. Bureau of Economic Analysis) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

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.

For this exercise. We are given the following information. So we have these hypotheses and we're told that the true parameter P is p prime and P prime is less than the null hypothesis proportion. So therefore the alternative hypothesis is actually true now. For part, they were asked to show the following. Where's Ed is the test statistic for a one proportion test. So the one proportion test statistic is given as follows. Now, if we calculate the expectation on Zed, everything inside the brackets is constant except for the sample proportion. So the hypothesized proportion and and are both constant. So the expected value can be rewritten like this. This is because the expected value for a constant is that constant. Now, we also know from Chapter two that the expected value for a sample proportion is the true value. So this becomes and this is what we're trying to show, at least at least with respect to the expected value. So now, for the variants now, in our test statistic again, everything is a constant except for this sample proportion. So that is the only varying parameter. Now I will rewrite this like this, so I'm taking this constant here outside of the variance brackets. So therefore I squared. So that's why these square root symbol has disappeared. And we also know from Chapter two that the variance for a sample proportion is given by the following. So therefore we have the following result and this is what else we were trying to show in part A. So that completes part A. Now in part B, we want to show at the power of the lower tailed test is that is we want to find the probability of getting a test statistic lesson or equal to negative of the critical value when the alternative hypothesis is actually true. Now, if our test statistic is that and we have just found its expected value and it's variants, we would expect that this value is normally distributed as standard normally distributed because we are taking a parameter, we're subtracting its mean from it and then normalizing by its standard deviation. So to find the power of the test, we're looking for the probability that the standard normal distributed parameter is less than negative, the critical value and now just plugging in our values for the standard deviation, which is the square root of the variance of the test statistic and the expected value for the test statistic. And if you look at what's asked for in question be when you were asked to show this right here now for Part C were given this information. So we have some hypotheses and were given the true proportion as 0.8 and we're testing at a significance level of 0.1 This means that are critical. Value is negative, 2.3 to 6, and we are also given n equals 225. And so we know that we know that the alternative hypothesis is actually true and were asked, What is the probability that our test will detect this? What is the probability that our test will result in us rejecting the null hypothesis? In other words, what is the probability of us getting a test statistic less than minus 2.3 to 6? So all we have to do here is plug in all the numbers into this formula, and this gives us a value of 0.978 So the power of this test, the probability of rejecting the null hypothesis when the alternative hypothesis is true, is 0.978

Okay, so we are conducting a one tale test at the 1% significance level Arnold hypothesis is that they're the same. And the alternative is that the first is less than the second the computer have statistic is going to be s one squared divided by S two squared. So this turns out to be approximately 0.17 Our degrees of freedom are one less than each of the sample size is making 18 the numerator and 50 the denominator, We can go to the back of the book in table eight to figure out what are critical value is. Since this is a left tail test, our f statistic is going to be f of one minus point to one, which is going to be equal to F of 099 which has a statistic of 2.78 Of course we take the reciprocal of that. This turns out to be approximately 0.36 So for illustration purposes, we have our F curve, whatever it looks like, and 0.36 is going to be somewhere right here, and 0.17 is going to be clearly on this side in the rejection region. Therefore, we reject the null hypothesis at the 1% significance level.


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