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7.2.35-T Tutoring Question Help 0 The pulse rates of 148 randomly selected adult males vary from a low %f 36 bpm t0 a high of 104 bpm Find the minimum sample size r...

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7.2.35-T Tutoring Question Help 0 The pulse rates of 148 randomly selected adult males vary from a low %f 36 bpm t0 a high of 104 bpm Find the minimum sample size required to estimate the mean pulse rate of adult malesAssume that we want 99% confidence that the sample mean is within 3 bpm of the population mean. Complete parts (a) through (c) below a. Find the sample size using the range rule of thumb to estimate &.n= 214 (Round up to the nearest whole number as needed ) b. Assume that o = 1

7.2.35-T Tutoring Question Help 0 The pulse rates of 148 randomly selected adult males vary from a low %f 36 bpm t0 a high of 104 bpm Find the minimum sample size required to estimate the mean pulse rate of adult malesAssume that we want 99% confidence that the sample mean is within 3 bpm of the population mean. Complete parts (a) through (c) below a. Find the sample size using the range rule of thumb to estimate &. n= 214 (Round up to the nearest whole number as needed ) b. Assume that o = 11.8 bpm; based on the value $ = 11.8 bpm from the sample 0f 148 male pulse rates n= (Round up to the nearest whole number as needed ) 'nt es9



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?

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.

Okay. Eso part a asked us to look at the box plot on DSO Looking at the box buff number 16. It wants us to make a conclusion about the differences between male and female pulse rates. And it looks like, um, like, the average in the median because they both look about normally distributed are about the same. And so, uh, my conclusion would be that, um, the differences in the means, uh, it would be, ah, close to zero. So my hypothesis would be they're close to zero S o. B. That's just to make a 90% confidence interval. So I'm gonna use the formulas found in the text. Um, that's basically the difference is so excellent minus X to bar plus minus my T score times the square root of my standard deviation of my first set squared, divided by n plus my standard deviation My second set squared about about in Oh, I'm about to put it back in can. And then that was for in one into. Okay, so, um, I can't call him a t score based off of, um, the degrees of freedom from adding the two total total to get 52 then subtracting 21 from each of them. So used degrees of freedom of 50 playing over. And I got a confidence interval that ranged from negative three point 011 And it went Teoh positive. 3.261 And that was for ah, degrees freedom of 15 which was 28 plus 24 minus two. Everything else he could find from the chart that I gave you above, uh, hurt. See, Does the compensated will confirm my answer part A. I believe it does because we were looking for the difference to be zero appear. Um, so I would say yes because zero falls on our 90% confidence interval. Okay, so it is within that interval. Thank you very much.

Let us look at this question. We want to find the sample size that is required to estimate the population. Mean we're using the body data. He does it one in appendix B, which includes weights off 1 53 Randomly selected males on those weights have a standard deviation off 17.65 So the standard deviation in this case is 17.65 17.65 This is my s. Okay, Because it is reasonable to assume that weights off mill statistics students have less variation than the weights of the population of 100 males. Let's Sigma is equal to 17.65 All right, so we are going to use Sigma is equal to 17.65 in this case, How many will statistics students must be weighed In order to estimate the mean weight off all male statistics, students assume that we want 90% confidence and the assumed that we were 90% confidence that the sample mean is within 1.5 cases of the population mean Okay, so we want 1.5. Casey has our margin of error. 1.5 is equal to Zed Alfa by two into Sigma by route n How many statistics students do we need? Gonna find the value of n and this is also equal to Sigma. And what is the value of the alphabet? First of all, what is Alfa Alfa is going to be 0.1, Scialfa by 20.5 In that case, what is going to be the critical value off or the Zen statistic that we use in this case? It is going to be 1.644 So this is going to be 1.6441 point 6449 Okay, Now let us use a calculator in order to calculate these values. So this is going to be 17 point 65 multiplied by 1.6449 divided by 1.5 and I have to square this. So this is 3 74.61 which means 3 75 and is equal to 3. 75 and is equal to 3. 75. These many number off statistics students, statistics male students must be weighed in order to estimate the mean weight of all male statistics students. Okay, does it seem reasonable to assume that weights of all weights off male statistics students have less variation than weights off population off adult males. No, it does not seem reasonable in this case, and this is my answer.

Now let us look at this question. This is what we did in the last question. Let us read this one again. We're using the desert one more editor in appendix B, which includes 1 40 cent randomly selected adult females. And those ages have a standard deviation of 17.7. And here they're saying we able to assume Sigma also equal to 17.7. So in this case, this value is going to be 17.7 17.7. Okay. How many female statistics students ages must be obtained in order to estimate the mean age of all female statistics students. Okay, we want 95% confidence and that the sample mean it's within 1.5 years off. The population mean it's me. This is correct and we want 95% confidence this time. So this value of Z Alfa by two is going to change. What is going to be Alfa by two? It is going to be 0.0 to five and alphabet to Z Alpha by two is going to be 1.96 This value turns out to be 1.96 in this case 1.96 1.96 I just have to substitute the values I have Sigma I have zf of I do and I want to find out the value off n So the value of end is going to change here. And after doing this calculation, the value off and turns out to be 4815 and is equal to and is equal to four eat one fight. My N turns out to be 4815 And the next question is, does it seem reasonable to assume the ages of female statistics? Students have less variation? Yeah, you can say that the sample size is pretty large here, and we Can anybody see that by looking at this analysis?


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