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Claim: The mean pulse rate (in beats per minute) ofadult males is equal to 69.2 bpm. For a random sample of 125adult males, the mean pulse rate is 69.5 bpm and the ...

Question

Claim: The mean pulse rate (in beats per minute) ofadult males is equal to 69.2 bpm. For a random sample of 125adult males, the mean pulse rate is 69.5 bpm and the standarddeviation is 11.3 bpm. Complete parts (a) and (b)below.a. Express the original claim in symbolic form.(1) _______(2) __________ bpm(Type an integer or a decimal. Do not round.)b. Identify the null and alternative hypotheses.H0:(3) _________(4) ___________ bpmH1:(5) _________(6) _____________ bpm(Type integers or decimals. Do n

Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69.2 bpm. For a random sample of 125 adult males, the mean pulse rate is 69.5 bpm and the standard deviation is 11.3 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form. (1) _______(2) __________ bpm (Type an integer or a decimal. Do not round.) b. Identify the null and alternative hypotheses. H0:(3) _________(4) ___________ bpm H1:(5) _________(6) _____________ bpm (Type integers or decimals. Do not round.) (1) μ σ p (2) = ≠ < > (3) p μ σ (4) ≠ < = > (5) σ p μ (6) > ≠ <



Answers

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want $99 \%$ confidence that the sample mean is within 2 bpm of the population mean. a. Find the sample size using the range rule of thumb to estimate $\sigma .$ b. Assume that $\sigma=11.3$ bpm, based on the value of $s=11.3$ bpm for the sample of 153 male pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to be better?

In this case, the real name is that the mean pulse read in the adult males is 72 on the P value that we get 010095 0.95 This is my pea Veloute. My Alfa is 0.5 So since the a P values less than Alfa, I will reject the null hypothesis. Now, what was the null hypothesis? The claim is that the mean, mean pulse rate is 72. Okay, so this was the original claim, which means this will be my null hypothesis. Final hypothesis will be so remind. Null hypothesis will be that my mu is equal to 72. And my alternative hypothesis will be that my mu is not equal to 72. Can I write it like this? Yes, I think I can. And I can say that I have enough sufficient sufficient evidence to reject the claim that the mean pulse rate off the adult males is 72 beats per minute. Hence I reject my null hypothesis. The original flame can be rejected

The original name in this case is that the standard deviation off pulse rates of adult males is more than 11 beats per minute. Okay, so this is the claim. So in this case, I can say in this case, I can say that my this is the claim, right? So this is my alternative hypothesis. My alternative hypothesis is that this standard deviation is more than 11. So this is greater than 11. And my null hypothesis will be that this is equal to 11 right now, at Alpha's equals 0.5 my P value is 0.3. And since my P value is greater than Alfa, this I get as 0.3. What is the 0.30453045? Now, since this is greater than my Alfa off zero point 05 I will say that I do not have enough statistical evidence to reject my little hypothesis. Right. So I do not have enough statistical evidence to claim that the standard deviation off pulse rates of what it means is more than 11. This would be my answer

Let us look at this question. I have 104 and 36. So 104 minus 36 is going to be my range. This turns out to be 68 now 68 is my range. Now I can assume my standard deviation to be 68 divided by four which turns out to be in this case by the standard deviation turns out to be close to 17. This is one seven. This is equal to 17. This is 17. So I have my sigma. This is going to be my Sigma. And will you treat this as my standard deviation? Okay, now I want in 99% confidence in trouble, which means my Alfa by two is 0.5 and Z Alfa by two is going to be 2.575 So 2.575 This is 17 upon Route N multiplied by This is multiplied by 2.575 and this is actually going to be equal toe to this is equal toe to right. So what is my end? My end in this case turns out to be 480 for a peek. All right. Now, the second one, we have already done the first one. Now assume that Sigma is equal about 15 based on the value of X is equal to 15 For example, off 1 47 female. Both states compared the results from party and be with the result is likely to be better. Okay, so this time the only changes. This is going to be 12.5. This is going to be 12.5 12 point faith. And this calculation turns out to be 12.5 to 2.575 divided by two. And this is raised to this is to 59 0.0 It's so I have to consider and to be to 60 in this case. Okay, so let me just check this calculation ones. Is this turning out to be 2 16? Yes. This end is turning out to be 2 60 now. My next question is which off? These is better. Which result is likely to be better. And the first one we are considering standard deviation as two in the second one. We are considering standard deviation as 12.5 There is more spreading 12.5. So we are leaving a more way. Are leaving more room to wiggle out if we make a mistake. So I think that Sigma is equal to 12.5 and this result is likely to be better. The result when I use signals equals five is likely to be better. Why? Because this time we're considering that the standard deviation is more when were considered standard deviation is to the graphic will look something like this. So yeah, the second one is better.

We have to random variables in this exercise, so the first one has a mean of 70 and a standard deviation of 10, and the second one has a mean of 77 standard deviation of 12. X bar is the sample average for a sample of 40 from the first population, and X y bar is a sample average for a sample of size 36 from the second population. Now for part A, we were asked for the approximate distributions of the sample averages. So for X bar, this is equal to the expected value of X, which we know 70 and the standard deviation of the sample average is the standard deviation of X, divided by the square it of them that comes out to 1.58 and then we could do the same thing for why so those air the means and standard deviations of sample averages now, in terms of the shape for the first sample average, we have sample size of 40. So therefore, by the central limit, their um we can say that the sample average is approximately normally distributed and we can say the same thing for the second sample average since it has the size of 36 and next for Part B, were asked for the approximate distribution of X bar minus y bar. So let's begin with the the means of the expected value of X bar minus white bar. Due to the linearity of expectation, we can write this as the expected value of X bar, minus the expected value of Y bar, which gives us minus seven and now for the variance. So assuming X bar and Y bar are independent samples, then we can say that the variance of X bar minus y bar is equal to the variance of X bar, plus the variance of y bar. And that comes out to 22. And from per eighth we found that X bar and why bar are both approximately normal, So therefore, their difference is also approximately normally distributed. So we could say that X bar minus y bar is approximately normally distributed with a mean of minus seven and a standard deviation of 22. Next for part C, we look for the probability that X bar minus y bar is between minus two and minus and plus one, and that comes out to 0.52 So the probability that expire minus y bar is between minus two and one inclusive is approximately 0.52 and this should really be approximation near, since we're approximating it with the normal distribution. And for part D, we're looking for the probability that X bar minus y bar is that most minus 15. And so we find that probability to be approximately 0.34 And we're also asked if we had actually observed a sample mean difference of minus 15. Would we doubt that the true difference in means is actually minus seven? So we just found that we have the probability of 0.34 of expire minus y bar being at most minus 15. So we have we stand a very good chance, actually, of finding that expire. Minus y bar is less than or equal to minus 15. We might also note that minus 15 is less than half a standard deviation away from the mean. So it's it's not very unusual to be about half a half a standard deviation away from the main, so we could say something like this


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