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3. In 1990,the mean height of women 20 years of age or older was 63. inches based on data obtaired from the Centers for Disease Control: Suppose that random sample ...

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3. In 1990,the mean height of women 20 years of age or older was 63. inches based on data obtaired from the Centers for Disease Control: Suppose that random sample of 45 women who are 20 years of age or older today results mean height of 63 inches with standard deviation of 3.46_ Conduct hypothesis test t0 assess whether women are taller today. Use 0.10 level of significance_For your hypothesis test; be sure State the null and alternative hypothesis State the type oftest: lefi-tailed. right-tail

3. In 1990,the mean height of women 20 years of age or older was 63. inches based on data obtaired from the Centers for Disease Control: Suppose that random sample of 45 women who are 20 years of age or older today results mean height of 63 inches with standard deviation of 3.46_ Conduct hypothesis test t0 assess whether women are taller today. Use 0.10 level of significance_ For your hypothesis test; be sure State the null and alternative hypothesis State the type oftest: lefi-tailed. right-tailed or tWo-tailed, check requirements Find and state the critical valuels) Use the formula find the test statistic. Show your work Calculate the p-value using tedf: Show your work: Draw picture; labeling the critical value(s), the rejection region and a, the test statistic and the p-value Write your initial conclusions using the classical method and the p-value method Write formal conclusion in context



Answers

In $1990,$ the mean height of women 20 years of age or older was 63.7 inches based on data obtained from the Centers for Disease Control and Prevention's Advance Data Report, No. 347. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 63.9 inches. (a) State the appropriate null and alternative hypotheses to assess whether women are taller today. (b) Suppose the $P$ -value for this test is $0.35 .$ Explain what this value represents. (c) Write a conclusion for this hypothesis test assuming an $\alpha=0.10$ level of significance.

However, one. So today we're doing a little bit more work on this state asset, which is about the height off women and specifically on female health professionals in America. And we did a little bit of work on confident falls in the last video using this data. And so now we're gonna be looking at testing the claim. That's the height of females in health. Fashion is different from the average height for women in the U. S. And so, just to remind you for a second off the values from this so this contains 50 values is contains 50 data points, and if you have them up on divide by 50 you're gonna find that the average is 64.7 eight. And we're told that the average height for females in the United States, which I'm gonna use me for is 63.7 inches, and the standard deviation of that is 2.75 Andi. So that should be all the information that we need to get working on. So to start with, we're gonna be testing a claim. So the first thing we need to do Yes. Oh, analyze first thing we need to do is we need to construct some hypotheses. So our hypothesis is that the average height off women, the health fashion is exactly the same is the average height of women in America. I said, that is 63.7. But I also I What's this is suggesting that the height of females in health freshen is different. And so that is saying that it is not equal to 63.7 on. This is important because this tells us we're doing a two title test. Yeah, And so next thing I want to do, let's calculate as itself all you So that star in a six X for which we have on the top left there minus mu all over Sigma, which is 2.75 I haven't reached 50. And if you put that into your calculator, you're gonna find that that comes out is 2.777 And so from this, we can find a P value. And so school just want sorry, really quickly. I've realized that I'm not quite clear on my notes whether I've were in the right thing down. So I'm using my country, you know, So we find out people using a calculator or using, um a the tables on you find that the p value for this I that stall value is, uh, no point. Not not 274 free. And so that's a p value on we're told to use the significance off No point nor five. However, remember earlier when I said it's really important that we know that this is two tailed. So that means we're testing both the upper and lower and and so what we can do we consult, approach this two ways, really? We can either say All right, Well, I want Alfa is gonna be no 20.0 to 5 on both of these ends. And then we compare this p value with no point nor to five. And what we find there is that this P value is significantly smaller. That so p is less than Alfa. So you reject ation all or alternatively, what you could do is you can say Okay, well, that's terrible, Curve. Let's try that again, is we can say OK, well, we're looking at both ends and I know that the area of this half is equal to my p value. And so, if I double my p value, I'm gonna get, um, not point. No, not ah. 5445 And then I compare that with the original Alfa, which is not quite nor five again. This should so show the exact same result, which is peace us an Alfa. So we reject. Okay, let me just make sure that that's enough. Enough for. And so that's two different ways of approaching a to tell. I thought sis, often I find this way easier because having what is usually a simple number isn't too difficult. Whereas doubling a longer decimal is a little bit harder. I suppose so. Either way, you're gonna get the exact same answer. And so what we find using this significance level is that we rejecting Haitian or so we're saying there's sufficient evidence using this sample that the high off females and health fashion is different from the average height of all females in the US So next part is a wordy question. It's suggest it's asking us how the rejection of the null hypothesis is revealed in the confident for that we came up with in the last video. So I'll just remind you of what that waas now. So the confidence before we came up with is this 65.54 So that's the continental we came up with on just to remind you again that the average height of women 63.7 sort of keep flicking up. And so we asked how we can use this confident fall to find that we would reject the null hypothesis. And essentially, what we said in the last video is that this value isn't in this interval on what that tells us is that we're certain within a 95% within about 95% that and we would reject the null hypothesis as they were actually doing the exact same thing by constructing a 95% confident interval were doing on then seeing whether this value is in that interval. We're doing the same thing if a little backwards to using a five point not five, um, level of significance in this test and then testing it that way. So the 95% confident Feliz the same is testing a 5% confidence in a test. I hope that makes some sense and specifically to reiterate the way that this reveals a rejection or hypothesis is because it does not contain this value that were given earthy national average. Then we're asked to explain how this interval could have been used to test the claim. Andi, that's exactly what I've just been saying. So we can construct a confidence interval like we did in the last video. And then we can essentially test whether the value that we're using in our hypotheses that we've got here is in the internal. And so what again that is essentially doing is exactly the same as this test. Just a little bit backwards. We would call this set of values the critical region. I'm not sure if this is technology you've come across before critical region, and what the critical region is is essentially the values for which you would reject. You would, um, reject the null hypothesis. And so not in this case is Excell Essence in before 0.2 four x greater than 65 0.5 full. And so those are the values that we would reject each not for on. So this value is in this critical region because it's less than 64.2 And so we reject a channel. And that's how we can use conference intervals to test the mean. And the final question is asking, How would we find a situation where we didn't reject the null hypothesis using a calm presentable and we wouldn't reject a the no hypothesis if the average was inside of this interval? If the so, if the actual average was between these two values, you know, if the population average, it would comparing again 64.2 then we wouldn't reject. And so that's how you would reveal no significant difference using a confidence interval.

All right. In this question, we wanted to test the hypothesis that new equals 63.7 Against the alternative hypothesis me greater than 63.7. For an n equals 45. X four equals 63.9. With the population of the inter deviation sigma equals 3.5 known at an alpha equals 0.5 confidence level. This question is testing our understanding of how to conduct hypothesis tests for population mean new with sigma known. So we proceeded this 23 steps listed first. We calculate the test net zero equals x minus mu over sigma. Right? Or you can remember that United 63.7. So, plugging in our numbers, we have zero equals 00.383 at alpha equals 0.5 significance. We can complete the critical value as 1.6 point five. This is from a Z table where the area to the right of zC on a normal curve is 50.5 Thus we conclude we reject H not. We failed to reject H. Not because then it is not in the critical region, are Xenon is the left of RCC, therefore it's not in the critical region.

So we're trying to find out if we have evidence to show that supermodels actually have a height that is higher than is the main hyper super rattles higher than the mean of 162 centimeters and you look at the data and it looks pretty evident that it's going to be higher. So I took my data and we have 16 pieces of data, put him in my calculator into a list and found that that mean value came out to be 177 0.223 and that's quite a bit larger than 1 62 especially when we look at the standard deviation being 1.844 So it's quite a bit quite a bit higher for a mean and the variability it's pretty pretty tight variability for that group. So we'll assume that supermodels actually have a mean height of 162 just like the normal population of women. And we want to find how likely it is to have a mean of 16 supermodels having a randomly having this and this value will be R. P. Value. And we'll find our test statistic. So if the mean is 1 62 how likely is it to get a group of 16 women and find that their height is greater than or equal 277.25 So let's change that to our test statistic which will be a T. Value with 15 degrees of freedom. And we have 1 77.23 Or excuse me to five minus the mean, we're assuming divided by the standard deviation from the sample over the square root of an and when we get that statistic we find out that that T. Value is absolutely huge, 33.8 and using my software with my T. CDF, I plug that in and found my P. Value and it is 9.8 times 10 to the negative 16th power. It is almost zero. Just a tad bit bigger than zero. So I don't care what significance level you use. We're using a 1% significance level and this is definitely less than 1%. So we have strong evidence to reject the nal mm and we can conclude that's super. Models do have heights greater than mhm. 162 centimeters. We aren't going to claim that there are 100 and 77 but we're just going to say it's bigger than 1 62.

Right. We have a data sample with n equals 38 r equals 21 corresponding to 21 correct options out of 38 corresponding to a certain problem we want to solve we want to test the claim that P for the population proportion is less than 0.67 with a significant level of 5%. This corresponds to a confidence level alpha equals 0.5 In order to conduct this hypothesis test will go through the following procedural steps following the identification of the confidence level just completed first. Is it appropriate to use the normal distribution to solve this? Yes. Both M P and Q P A greater than five. What are the hypotheses hypothesis is that H not P equals 50.67 H a p is less than 0.67 Ak We're using a one tailed test compute P hat and the test statistic P hat is all over. End 10.55 plugging that into RZ stat formula on the right yields negative 1.57 for Z. Next we use the computer P value. We use a Z table from which we find the P value is 0.582 That is the area under the normal curve left of our Z stat, as highlighted here on the right. Finally, we reject H not No, we do not because the p value is greater than alpha, and we interpret that to mean that we lack evidence that P is less than .67.


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