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QUESTION 12Explain wha:the P-value Miesne forthis Prod Jem; If Ho is false, then there chance qual t0 thevalue that the value of the test statistic will be equal to...

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QUESTION 12Explain wha:the P-value Miesne forthis Prod Jem; If Ho is false, then there chance qual t0 thevalue that the value of the test statistic will be equal to or greater than the calculated value_If Ho is false_ theh there is chance equal to the p= value that the value of the test statistic will be equal t0 or ess than the calculated value If Ho is true, then there chance equa to the p-value that the value of the test statistic will be equal to or less than the calculated valueIf Ho is tru

QUESTION 12 Explain wha:the P-value Miesne forthis Prod Jem; If Ho is false, then there chance qual t0 the value that the value of the test statistic will be equal to or greater than the calculated value_ If Ho is false_ theh there is chance equal to the p= value that the value of the test statistic will be equal t0 or ess than the calculated value If Ho is true, then there chance equa to the p-value that the value of the test statistic will be equal to or less than the calculated value If Ho is true theh there is chance equal t0 the p- value that the value of the test statistic will be equal to or greater than the calculated value_ QUESTION 13 Decision: Should the null be rejected? Enter either yes or no QUESTION 14 Reason for Decision: Enter either Or > p-value QUESTION 15 Conclusion: There sufficient evicence Concuo that there Gaterence 2mono different nutritiona Tortwlas rats with respect weigh: gain_ There sufficient evidence conclude that there difference mong the different nutritionz fonulas for rats wich respect weigh: gain.



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(i) What is the level of significance? State the null and alternate hypotheses. (ii) Check Requirements What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding $z$ or $t$ value as appropriate. (iii) Find (or estimate) the $P$ -value. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (v) Interpret your conclusion in the context of the application.Note: For degrees of freedom $d . f$. not in the Student's $t$ table, use the closest $d . f$. that is smaller. In some situations, this choice of $d . f .$ may increase the $P$ -value a small amount, and therefore produce a slightly more "conservative" answer.Answers may vary due to rounding. wanagement: Intimidators and Stressors This problem is based on information regarding productivity in leading Silicon Valley companies (see reference in Problem 21 ). In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him-or herself talk. Let $x_{1}$ be a random variable representing productive hours per week lost by peer employees of an intimidator.$$\begin{array}{llllllll}x_{1}: & 8 & 3 & 6 & 2 & 2 & 5 & 2\end{array}$$. A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let $x_{2}$ be a random variable representing productive hours per week lost by peer employees of a stressor.Use a calculator with mean and standard deviation keys to verify that $\bar{x}_{1}=4.00, s_{1} \approx 2.38$ $\bar{x}_{1}=5.5,$ and $s_{2} \approx 2.78$ (a) Assuming that the variables $x_{1}$ and $x_{2}$ are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a $5 \%$ level of significance. (Assume that the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetrical.) (b) Find a $90 \%$ confidence interval for $\mu_{1}-\mu_{2}$. Explain the meaning of the confidence interval in the context of the problem.

The following is a solution to number 23 and this is about rabies in in Fox. Uh Fox regions. And the first part is just to verify that the sample mean for region one was 4.75 and that standard deviations 2.82 And then the sample, I mean for the second region is 3.93 and the standard deviation is 2.43. And they say use a calculator. So I'm assuming it's going to be a some sort of scientific or graphing calculator. So I went ahead and pre populated. If you go to status is a c. and if you go to stat edit, I went ahead and punched in those numbers already into L one, L two and then you can find these values fairly quickly. If you just go to stat and then one of our stats and then do your L one. So this is the verification. So X bar is 4.75 which what it was and then s. is about 2.82 and then would verify on the second one if you go to stat couch and then it's one of our stats and then second L. Two And we calculate that and that's verified. So 3.93 and then two 43 for the standard deviations, that's verify that first part Um using some technology, so that's how you would use it in the IT4. And the second part of this we actually do a um hypothesis test without formally writing everything down. So this is a two sample T test since we don't know what the population standard deviation is, just the sample standard deviation. So we're gonna use a T. Test and the significance levels point oh five. And what we're gonna do since we already have the data there, let's go ahead and say stat and then test and it's the two sample T. Test and I'm gonna keep it on data here because I already have the data listed into L. One L. Two and then the everything else can say the same. But the alternative um it says that that the two regions are just different either way so we don't know which one is bigger which was smaller. So whenever that's the case you're just gonna leave it as not equal to. And generally we're going to keep keep it as no on pooled unless it tells you otherwise. And then we calculate and that's gonna give us what we need. So the T. Value is 0.865 But really all I care about is this p value 0.394 So the p value so m. One is not equal to end to. That's the the alternative I guess and then the Noles that they're the same. But the p value is the important things of the P value equals point 394 And what we do is we explicitly compare the P value with the alpha value and in this case the p value is greater than alpha. So that means we fail to reject h not meaning we're not rejecting, meaning that these two means are the same. So if we put it and um you know, normal words, writing what we can say, I'm gonna type this because it'll be faster. There is not sufficient evidence to suggest that the mean number of cases of fox rabies is different In the two regions. So it's basically just saying that um these the mean number of rabies in these two regions are the mean number is approximately the same.

We want to conduct a pair differences test at alpha equals 5% significance testing. The claim that the population means X. A and X. B are not equal. The data is given below. With mount shapes and at your distribution on the right. I've already computed D. Bar 2.25 The sample size N equals eight and the standard deviation S. D equals 7.78 We proceeded to five steps listed below to solve first. We'll check the requirements and evaluate hypotheses. So because the distribution shape requires have been met to use the student's T distribution, the degree of freedom is at minus 27. We have no hypothesis. H. And R equals zero alternative mu D. Does not equal zero Alpha equals 00.5 significance. Archer statistic is T equals D. Bar over SD over route and 4.8181 from the tea table R. P value is 0.5 is less than ph 7.25 So we have enough information to conclude our test. Now, since P is greater than Apple, we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal zero.

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

We're told the population has mean new equals 8.8. And are given in the following sample data randomly selected from that population from that sample thing that we want to calculate the mean X bar and the sample standard deviation S. So to do so let's remember the formulas for each of these terms, sample mean X bar is defined as the sum of the data that I am in this case 7.36 And the sample standard deviation is defined as some of the deviation about the mean squared all divided by n minus one. In this case, 4.3 Next. What we want to do Is used the sample data to implement a two tailed test. That is we want to test whether or not the sample data suggests the actual meaning. The population differs from the nomen of 8.8 in either direction, higher or lower. We want to use significant level Alpha equals 0.5 And we note at this point that X. Is approximately normally distributed. So to answer this test question, we have to go through the following procedures in order to finalize The solution of this test. So first, what's the significance level? And hypotheses are physical. 0.05. No hypothesis H. Not musicals. 8.8. H. A mute does not equal the population. 8.8. What distribution are using completely associated test statistic? The distribution. We're going to use the student's T distribution because our population standard deviation sigma is unknown. We can definitely use this distribution though, because the shape is appropriate. Ak because X is approximately normally distributed, it's okay to use a student's T distribution. Next, what is the T stat stat is defined by this formula, which here we see, reduces down to T equals negative 1.337 Next let's use the T stat to compute the p and dribble and sketch this out on the student's T distribution. So since we have degree of freedom and minus one equals 13, we're going to use the two tailed T table, which you can find either on google and a staff textbook to identify what range of p values this T stat negative 1.337 falls between, we find it falls between 0.2 and 0.25 We can graph this as the area underneath the student's T distribution, which we highlighted here in yellow outside of our T stat. Uh In this case we have both negative 1.337 and positive 1.337 Because the detailed test next, what can we conclude from this? Well, he is greater than alba. So we have statistically insignificant findings, i. E. We cannot reject h not or no hypothesis. And that ultimately means that we lack evidence suggesting that the population mean differs from its no mean of a pointing


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