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Researchor canteddeterminecarpeted rooms Contain more bactera than uncarpeted rooms The table shows the results for Ihe numberbacteria per cubic IootFull data s0 Un...

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Researchor canteddeterminecarpeted rooms Contain more bactera than uncarpeted rooms The table shows the results for Ihe numberbacteria per cubic IootFull data s0 Uncarpetedtor both types roomsCarpeted 11.9 12.3 12.9 12.8B. Ho: 4; = Hz H1:H # H2 Ho' H1 < H2 H; 0, > Hz Ho: H, "02 H,: H < HzDetermine the P-value for this hypothesis test P-value(Round three decimal places as needed )

researchor canted determine carpeted rooms Contain more bactera than uncarpeted rooms The table shows the results for Ihe number bacteria per cubic Ioot Full data s0 Uncarpeted tor both types rooms Carpeted 11.9 12.3 12.9 12.8 B. Ho: 4; = Hz H1:H # H2 Ho' H1 < H2 H; 0, > Hz Ho: H, "02 H,: H < Hz Determine the P-value for this hypothesis test P-value (Round three decimal places as needed )



Answers

Find the rejection region (for the standardized test statistic) for each hypothesis test. Identify the test as left-tailed, right-tailed, or two-tailed. a. $\quad H 0: \mu=141$ VS. $H a: \mu<141$ [email protected] \alpha=0.20 .$ b. $\quad H 0: \mu=-54$ VS. $H a: \mu<-54$ @ $\alpha=0.05 .$ C. $\quad H 0: \mu=98.6$ VS. $H a: \mu \neq 98.6$ [email protected] \alpha=0.05 .$ d. $\quad H 0: \mu=3.8$ VS. $H a: \mu>3.8$ @ $\alpha=0.001$

All right. So in this problem we are going to be testing a um test of hypotheses using the p value approach. So for part a the first thing we have to do is we have to calculate our Z score. Um And in order to do that we're going to take our observed proportion, subtract are expected proportion and divide by our standard deviation of our sampling distribution. And when we do that we get our Z score of 2.24. Now, in order to calculate our P value, this is a two tailed test. So we're going to find the probability of one tail using our um Using Our Appendix chart. So 2.24, this is from chart to To find our tail probability -22 four is 0.0125. And we are gonna multiply that by two. Since there's two tails which gives us Our P Value of zero 25 Which is greater than our significance level of .5. So in this case we will fail to reject r knoll. All right. That was a two tailed tests that we had to find our P value using our chart and then multiply it by two to include the other tales. Since the graph is symmetric, part B again we have to calculate our Z score. So 0.96 minus are observed are expected to 0.94 divided by the square root of the standard deviation of our sampling distribution Divided by our sample size of 1100 which um gives us our Z score Of 2.92. And this one is a one tailed test. We're just testing if it's greater than So here's our Z score and we got to figure out what that area under the curve is, The area under the curve is R. P. Value that we're going to compare out there. So using our chart, this one, since it's a right tailed test, we have to use our chart and then take the compliment of it, which we get our P value of 0.18, which is less than our alpha of point oh five. So we would reject the null here because it is low, so if the P is low we reject the ho. So in this case we would we would reject r knoll because it is less than point oh five.

All right. So we're still kind of looking at the rejection region for the standardized test statistic. Um But they also want us to clarify here what the uh if it's a left tailed right, tell her to tell which we've already kind of been looking at. So we're looking at the chance of um New being less than 141 with a significance level of .20. Since it is less than it's only a one tailed test. And that is to the left. Okay, basically the same as what I've drawn up here. We're looking for just this tail probability. So if this tail probability is less than .2 Um then we would have rooms for rejection. So then that means on our um for our standardized test statistic, I'm gonna look at the tail probability of .2 which is anything when the Z score less than or equal to negative point A 42 and it's negative because we're on the left hand side. Remember the Z score is centered at zero. So we're to the left of that. Since the left held test, we're looking for a Z score less than or equal to the negative Z score from the critical values of teach our all the way at the bottom. Alright, so part B we are testing Mu less than 54 And our significance level is 5%. Again less than it is going to be a left tailed test. And our answer's gonna look very similar to what we have before. Just with a different significance tests are significance level which is -1.645. Again, that's straight from figure 12.3 in the back of the book, as the Appendix For part C. We are testing μ not equal to 98.6 At a significance level of .05. That is not a left field test. That is a two tailed test. Because it could be greater than 98.6. Extreme Less than 98-6 extreme. Either one of those extreme values would satisfy. So since it's to tell we have to divide our significance level by two Which gives us two options Less than 1.96 for Z score or greater than positive 1 96 as our Z score. Last one for this problem Greater than 38 at a significance level of .001. That is a right tailed test. Because we're looking for our significance level being greater than I mean being greater than 38. So RZ score we're looking for a positive Z score all the way down here on the right. Um So we're not splitting it in half .001 All the way down our Z score chart at the infinite degrees of freedom. Z score greater than 3.09 would give us room to reject the null. Given those conditions

E want to check if these first two results are significant or not at given Alfa levels. So my first you statistic is 1.92 and degree of freedom is 44. In order to calculate the P value for the first one with P values 0.3 which is less than Alfa hence, I can say that the first one is significant. The second one mighty statistic is 3.41 degree of freedom was 44. My advice 0.1 So this is three point for one degree. Freedom is 44 Alfa 0.1 What is my P value? My P value. My P value is much less than 0.1 If it is lesser than that, it is definitely less than this. Hence, we can say that both of these results are significant at these Alfa levels. The third one, the third one I have been given tee off 44 is equal to 1.81 Now what will be the P value reported in one tail test and to tail test? First of all, let's understand what is a p value p value Is the probability off an event or something even rarer than that offering? If it is a mental test, what is a P value for this? This is 1.81 somewhere. The P value is going to be this region. But if But if this is a two tailed tests, I am going to have 1.81 here. And there is also a minus 1.8 When here, minus 1.81 here. And this is going to be the twice off it. So p value becomes Christ. So let me just try toe find the P value P is equal to 44 1.81 at 0.5 0105 years ago. The physical 40 40 years ago to 1.81 He is equal to 1.81 The last one I think I put I had in the second one, I had mistakenly put their 13.41 But you will find that even if I put 3.41 my people, it will be much less than that one. Much less than 0101 Okay, so 1.81 degree a few dollars. 44. And let's try a one tailed test. First one tail does to 0.38 Fight. If this were one tail test only this region is 0.3 Eat, face something. But now if this is a two tailed tests, it will get doubled because I want the area off both the sides. So there's going to be 0.7 something. Let's see. This is a two tailed test. This becomes 0.77 This becomes there's a brutal test. This become 0.77 So, yeah, P values less than 0.3. Value is less than 0.1 in both the cases. But if it were a one tailed test, P value would have been this. And if it were to tell test, the value would have been this

All right. So in this question we are looking at determining the rejection region. Um for the test statistic also what tail it is. So for the first one we're testing against μ not equal to negative 62 at a really small significance level of .005. Which in this case um since mu is not equal to that is going to be a two tailed test Which means that that 5% probability is contained within two tails. That means we need to divide it by two to find the position. Half of that rejection probability is here, half of it is here. Um So when we divide that into We get .0025 and scrolling down to the infinite degrees of freedom on our Figure 12 3 and the Appendix Less than -287 or Greater than or equal to positive 287. Okay. For part b You greater than 73. So one tell test At a small significance level of .001. It's um to the right tail because it's greater than that means we're gonna be on this side of our curve and we don't have to divide our significance level into because all of that rejection probability is contained in one tail. So .001 for the tail probability scroll all the way down Greater than or equal to 3.9. Part C Mu less than 1124. I'm sure you've noticed like this number really doesn't matter. It's really the sign that is the only thing we're interested in here, it will matter Once we start actually computing the test statistic, then that's when things will start to matter. Um Which is coming in number five for this section. So again we're looking at a one tailed test that is left tailed here and so it's gonna be less than or equal to the negative Z score. We're going to look for that 0.1 tell probability which is negative 3.9, same as what we had in part B. And finishing things up in part D. We are doing a two tailed test because we're looking at not equal to 12 for our main With the significance of .001. So it is two tailed because of the not equal to. So we have our negative test, we have to split that tail probability in 2.001, split into two as .0005. And the Z score at the bottom there is 3.291. So that wraps up how to determine the rejection region. And then as we move on we will start looking at what things look like when we are actually finding the Z score


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