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(5 points) Is college worth it? Among simple random sample of 309 American adults who do not have a four-year college degree and are not currently enrolled in schoo...

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(5 points) Is college worth it? Among simple random sample of 309 American adults who do not have a four-year college degree and are not currently enrolled in school, 133 said they decided not to go to college because they could not afford school:NOTE: While performing the calculations; do not used rounded values. For instance, when calculating a p-value from a test statistic, do not use a rounded value of the test statistic to calculate the p-value. Preserve all the decimal places at each step:

(5 points) Is college worth it? Among simple random sample of 309 American adults who do not have a four-year college degree and are not currently enrolled in school, 133 said they decided not to go to college because they could not afford school: NOTE: While performing the calculations; do not used rounded values. For instance, when calculating a p-value from a test statistic, do not use a rounded value of the test statistic to calculate the p-value. Preserve all the decimal places at each step: Enter at least 4 decimal places for each answer in WeBWorK: 1. A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence_ What are the correct hypotheses for conducting a hypothesis test to determine if these data provide strong evidence supporting this statement? A. Ho p =0.5, Ha : p # 0.5 B. Ho p =0.5, Ha : p > 0.5 C. Ho p = 0.5, HA : p < 0.5 2. Calculate the test statistic for this hypothesis test: 3. Calculate the p-value for this hypothesis test_ 4. Based on the p-value; we have: A. little evidence B: strong evidence C. very strong evidence D. some evidence E. extremely strong evidence that the null model is not a good fit for our observed data_



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Parts (a) and (b) relate to testing $\rho$. Part (c) requests the value of $S_{e} .$ Parts (d) and (e) relate to confidence intervals for prediction. Parts (f) and (g) relate to testing $\beta$ and finding confidence intervals for $\beta$. Answers may vary due to rounding. Physiology: Oxygen Aviation and high-altitude physiology is a specialty in the study of medicine. Let $x=$ partial pressure of oxygen in the alveoli (air cells in the lungs) when breathing naturally available air. Let $y=$ partial pressure when breathing pure oxygen. The $(x, y)$ data pairs correspond to elevations from 10,000 feet to 30,000 feet in 5000 -foot intervals for a random sample of volunteers. Although the medical data were collected using airplanes, they apply equally well to Mt. Everest climbers (summit 29,028 feet). $$ \begin{array}{l|rrrrr} \hline x & 6.7 & 5.1 & 4.2 & 3.3 & 2.1 \text { (units: } \mathrm{mm} \mathrm{Hg} / 10) \\ \hline y & 43.6 & 32.9 & 26.2 & 16.2 & 13.9 \text { (units: } \mathrm{mm} \mathrm{Hg} / 10 \text { ) } \\ \hline \end{array} $$ (Based on information taken from Medical Physiology by A. C. Guyton, M.D.) (a) Verify that $\Sigma x=21.4, \Sigma y=132.8, \Sigma x^{2}=103.84, \Sigma y^{2}=4125.46$, $\Sigma x y=652.6$, and $r \approx 0.984$. (b) Use a $1 \%$ level of significance to test the claim that $\rho>0$. (c) Verify that $S_{e} \approx 2.5319, a \approx-2.869$, and $b \approx 6.876$. (d) Find the predicted pressure when breathing pure oxygen if the pressure from breathing available air is $x=4.0$. (e) Find a $90 \%$ confidence interval for $y$ when $x=4.0$. (f) Use a $1 \%$ level of significance to test the claim that $\beta>0$. (g) Find a $95 \%$ confidence interval for $\beta$ and interpret its meaning.

The following is a solution in over 14 goodness of fit test where we look at the distribution of customer ages in a snoop report and see if it gives it agrees with the report from the sample. So the observed data values here uh were given to you in the book. So 88 1 35 50 to 40 76 1 28. I went ahead and just copy those down. And then for the expected, you gotta take the percent Times the sample size to get the actual expected value. So 12% times 519, for example, is 62.28. So I did that for all six categories there for the different age ranges. So now we're ready to go ahead and To answer these questions. So what's the significance level that they give? They give you that in the problem? It's a 1% significance level. So alpha's .01. And then we also need to write down what are null and alternative hypotheses are now typically for goodness of fit, test the Noles that these two distributions are the same. And then the alternative is that they're not the same. So in this particular case, bringing it back to this example, we're going to say the distribution of customer ages, uh and the snoop report, I'm not sure what a stupid port is, agrees with the sample report. Right? And then the alternative is that they don't agree. So, the distribution of customer ages and the stoop report does not agree with those of the sample report. So now we need to find the chi square value, but before we do that, we need to make sure that the expected values are greater than five and they are. So the expected values already found them, and you just take the percent times the sample size and they're all greater than five. So that's one of those conditions for inference they need to be at least five and they are so we can use the chi square distribution because of that with five degrees of freedom. Now why is it that five degrees of freedom? Well because there are six categories 1234566 categories minus one is five. That's always gonna give you the degrees of freedom. So now we can find basically everything else with software. So I'm gonna use the T. IT. for like using the T. 84 for smaller data sets. Um But you can use Excel or are or SAs or mini tab whichever you prefer. Or you can certainly use the formula. It might take you a little longer but you can use the formula. So I went ahead and typed these in already but I put an L1 I put the observed data values so you can see him there and then an L. Two. I put the expected data values you can see him there. So what I like to do sometimes is look at the difference here and these look to be kind of far apart. At least some of them do. So I don't know. I think it's been close but I don't think this is going to be a very good fit. Well let's see. So if we're gonna stat and then tests and it's one of the last ones I went up first. It's the D option if you're using A T. I. T. For. And it's the chi square geo F. Test that stands for goodness of fit test. L one is your observed, assuming you put it in. L one. L two is the expected degrees of freedom. Was five. Remember? And then we calculate and that's gonna give us the two values that we need to. The chi square value. Is this first number right here? 15.651 15 651 And then the P. Value, I don't know if you saw it there but we'll go in round. It's about .008 0.8 So just what I thought it's less than alpha. So anytime the P values less than alpha we this is the beauty of the P value. Matthew explicitly compared the P value with the alpha value. And if it's less than alpha then you always reject the null hypothesis. So we're going to reject h. Not so we're rejecting the claim that the distribution of customer ages and the snoop report agrees with the sample report. So that's how we conclude that will go and say there is. Oops. Well, I didn't mean to do that. Yeah, I can just type it in I guess. Mm. Sorry. Okay. I'll just I'll just actually write this out so there is okay, sufficient evidence to suggest the distribution of customer ages does not agree with that of the sample report. Okay. So these two distributions are not the same.

So this problem has several parts to it. First thing I need to do is I need to identify the significance level and from the story problem at 0.01. My null hypothesis is that P is indeed 0.301. And then my alternate is that P is actually greater than We're told. We have a sample size of 228. R equals 92. and again the probability is 0.3 01. So that means Q is 0.699. This will be a standard normal distribution. Just checking his NP greater than five. And yes, it is because 228 times 0.301 is indeed greater than five because it is Approximately 68.6. And then when I test end times Q 2 28 times 0.699. that gives me an approximate value of 1594. And that is indeed greater than five. Now I need to find P hat and because we're doing standards normal, I'm looking for the Z value. I'm not going to do this by hand. I'm going to use technology so I've already got it done here. But let me work through it with you stat Test. This is a one proportion z test. So number five The probability piece of 0.301. My ex this goes with the r value that was 92. And my sample sizes 2 28. And we want to test greater than so I'm going to click on calculate and there's the information I need so I got this on the other screen so my P hat is approximately point 40 and then rounding let's say we'll go three places and then my Z value is 3.37 So if I'm drawing this on the curve Here's 3.37 and I'm shading to the right and then here's my P. Value. So for part C. I need to check is my P. Value less than greater than or equal to α. So this either the -4 means it's a very small number. That's a form of scientific notation. So zero. Then the six will make the three round up. So .0004. And is that greater than less than or equal to? It's less than. So that means we will reject the null hypothesis. And then what does that mean in this in context of the problem? So at the 1% significance The sample data indicates that the proportion of numbers in the revenue file with a leading digit of one exceeds zero 301 So at the 1% significance level the sample. So we reject them all. Okay so let's interpret that If p. is in fact larger than 0.301. What does that tell you? Does it seem that there are too many numbers in the file with leading ones? And the answer would be yes. I guess I could have take this one out too. Huh? It indicates There are too many numbers leading with one. So what does that mean? As far as I could this indicate that the books have been cooked, chances are you're not writing numbers that are too big. So it could be that there was an error somewhere. Um The IRS of course should investigate more because it is very unusual. Could be some kind of as it says in the book. Could be profit skimming. So they take The extra and then right in the books one. But the bottom line is the FBI should investigate. So it could be a mistake. Or it could be that somebody is actually writing lower numbers in than what are actually true. And then finally, what does it mean to reject the null in this situation? So it's really important to know that by rejecting the nol, we have haven't actually approved then all to be false. The data did lead us. So that indicates that too many numbers start with one. So more investigation is needed.

All right. We have a sample of size and equals 196. And in that sample we see 29 objects satisfying a certain observation we want to observe that is R equals 29 out of 10 equals 196. We want to use this data to test the claim that P is greater than 092 with alpha equals 0.5 Or confidence level 5%. Now that we've identified the confidence level, we can proceed in the following procedural steps in order to conduct this hypothesis test first. Is it appropriate to use the normal distribution? Yes, it is. Because N. P and Q. P. R both greater than five secondly what hypotheses are retesting? We're testing H and R P equals 50.92 H. A. P greater than 0.92 Which means we're conducting a right tailed or one tailed test. Next compute P. A. And the test statistic he had simply are over end or 0.148 plugging that as well as P. Q and N. Into rz stat formula on the right. Give Z equals 0.271 or 2.71 next let's compute the P value based on our Z equals 2.71 We can use this table to identify the P value. The P value is simply the area under the normal curve to the right of the Z score. Since this is the right tool test from the table, we get P equals 0.34 We've illustrated this in the graph on the right next. We use this P value to reject H. Not. Yes, we do because he is the alpha. And we interpret this to mean that we have evidence that P is greater than 0.92

All right. We have a sample of size and equal 73 of the 73 objects in the sample, 56 meet a certain criteria. And you want to observe our equals 56. We want to use this data to test the claim that the population proportion P does not equal 560.82 at a confidence level of 1% or alpha equals 10.1 Now that we've identified the confidence level, we can proceed in the following steps to conduct this test first is the normal distribution appropriate to use? Yes, it is because both NP and qpr greater than five. What hypotheses are retesting? We're testing them. No P equals 0.8 to the alternative. P does not equal 0.82 meaning we're testing or rather conducting a two tailed tests for this population proportion. Next compute p hat in your test statistic Piatt is simply are over and equals 20.77 We plug that along with PQ and and into the Z stat formula on the right to obtain Z equals negative 1.18 next to the P value. The p value can be found using it a lot google and textbook. And is the area outside of our positive, negative Z scores. This is a two tailed test, A. K. A. The area to the left of negative 1.18 An area to the right of 1.18 as is highlighted on the graph on the right. This gives P equals 0.2380 Next we reject H not no, we do not because he is greater than alpha. And we interpret this finding suggests we lack evidence that P does not equal 0.82


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