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Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes ages of 147 randomly selected adult females, and those ages have a standard deviation of 17.7 years. Assume that ages of female statistics students have less variation than ages of females in the general population, so let $\sigma=17.7$ years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want $95 \%$ confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that ages of female statistics students have less variation than ages of females in the general population?

We have the data set given below here and using it, we want to find the sample mean X. Bar, the sample standard deviation S. And constructed 99% confidence interval for the population we knew, assuming the population is normally distributed. So since we have this data we can find expire and s using the appropriate definitions expire, is that some of the data divided by n 2.353 and s. Is the square root of the sum of deviations of the mean square divided by n minus one. In this case 1.3 Now with X. Bar and sample standard deviation S, we can use a student's t distribution to construct this confidence interval. So we can identify the appropriate T score. So for the degree of freedom and minus one equals 14 and critical values um providing probability 140.99 for the confidence interval, we can use the students to table either on google or in a textbook to identify T C equals 2.977 Now, with this T. C, we can identify the margin of error E, which is given by this formula plugging in our T C. S and N. We obtain E equals 0.572 And now plugging into our confidence interval formula, which is that new falls between sample mean minus C, and sampling plus E gives us our confidence interval. That new is between 1.781 and 2.9 to 5 with 99% confidence.

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.

In question number 13. The population means is equal to 98 plus 95 plus 93 over ST what you call toe. 25 points. 3333. 30 street. Why the Population Standard Division is equal to the square with war 98 minus 25 points ST ST ST three square to 93 minus 95.3000. 333 square over three with approximately equal to 2.45 48. Note that the mean something distribution is equal to the population mean just under divisions are not equal. And at the end of this question, thank you.

So we are going to be examined. The heights of female students on the average of 64.4 m. And the standard deviation is 2.4 m. Just kidding. It's interest. Now, the C score is looking at x minus mu divided by sigma, and that's what we're doing to find. The first boundary of X. Z is 62. So we get approximately -1. Now, for the Z score of 63, We're just gonna replace the 62-63. Okay, But unfortunately I have to write this out again and it's lot to write out sometimes, but on a test you're gonna want to do that. And the difference that we find here Or a result is going to be -058. So what this tells us is that it is between Almost 1/2 and a whole standard deviation below the mean. And now to find out what the percentages are, we're just going to look in the back of the book here. Okay. Yeah. This one is going to be point oops, mm. Bad trying, bad trying. Okay, there we go. Now. We're back in order. This one is going to be 0.15 87 Yeah. And this other one is going to be a 15%,, A much higher. 0.281. So this is going to be our ends. Okay, looking at between here and here And what we have is between 10 And 28%. To find this difference right here, It's going to be point too late. one -1587. And this difference comes out to be 12.23%. Yeah. Oh, Finally we're asked to final percentage of female students with fights between 65 and 70 in so once again will be calculating see scores. Yeah. and I wont show that math. This time, it's just 65. It's going to be at 0.25 and for 75 are sorry 70 R. Z score turns out to be a whopping 2.33 For this one. Right here we get a percentage of the 99%ile .9901. And for this one you get 0.5871 Her 5987. I was looking at the wrong role. Yeah. Richard Zoom in or whatever. Trace your finger down just so you can get the right percentage. The percentage of female students are between these two heights is going to be the difference. The difference between these two is 0.3914 Or 39.14%. Okay.