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In & = snale of 27 MLS soccer games; the home team won 15 times: for home team winning proportion of 4 0.562 Uslng these datz bootstrap sampling distribution wa...

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In & = snale of 27 MLS soccer games; the home team won 15 times: for home team winning proportion of 4 0.562 Uslng these datz bootstrap sampling distribution was construcled in StatKey with - standard eror of 0.094.What Is the 95% confidence interval for the proportion of all MLS games won by the home team? Use the standard ertor method:[0.462.0 6501[0.415,0.697](0.368.0,744]I0.274,0 8381I0.180,0.932|6+0462.0. 650101*1,018_ 7081Questicn 6StatKey Is used t0 construct bootstrap sampling distri

In & = snale of 27 MLS soccer games; the home team won 15 times: for home team winning proportion of 4 0.562 Uslng these datz bootstrap sampling distribution was construcled in StatKey with - standard eror of 0.094. What Is the 95% confidence interval for the proportion of all MLS games won by the home team? Use the standard ertor method: [0.462.0 6501 [0.415,0.697] (0.368.0,744] I0.274,0 8381 I0.180,0.932| 6+0462.0. 6501 01*1,018_ 7081 Questicn 6 StatKey Is used t0 construct bootstrap sampling distribution t0 estimate populition correlation That distribution is approximately normal with mcan of 0.420 Jrd standard deviation 0f 0.096, Using the standard error method construct & 95% confidence interval t0 estimate lhe correlation in the corresponding population: I0.324,0 516] I0.228,0.6121 [0.132.0.708] [0.036.0.8041" 010.324.05161



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Baseball: Home Run Percentage The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional bascball players gave the following data for home run percentages. (a) Use a calculator with mean and standard deviation keys to verify that $\bar{x} \approx 2.29$ and $s \approx 1.40$ (b) Compute a $90 \%$ confidence interval for the population mean $\mu$ of home run percentages for all professional bascball players. Hint: If you use Table 6 of Appendix II, be sure to use the closest $d . f .$ that is smaller. (c) Compute a $99 \%$ confidence interval for the population mean $\mu$ of home run percentages for all professional baseball players. (d) Interpretation The home run percentages for three professional players are Tim Huelett, $2.5 \quad$ Herb Hunter, $2.0 \quad$ Jackie Jensen, 3.8 Examine your confidence intervals and describe how the home run percentages for these players compare to the population average. (e) Check Requirements In previous problems, we assumed the $x$ distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section 6.5.

The following is a solution # six. And this asks 785 people whether they follow college football. And of those 785 people, 275 of them said that they followed college football. So that was the prompt. And we're asked to find a 95% confidence interval in this scenario. So if you wanted to you could find the P. Hat. You just take the X. Over the end but you really don't need to. Especially if you're gonna use your calculus which I'm going to show you here in a second. But to 75 out of 7 85 um That's gonna be our P. Hat. And uh this is a proportion uh they in fact it's a one proportion Z interval. Okay? Because it doesn't ask anything about mean or anything like that. It just gives you a proportion. So that's what we're gonna do. So you can use the formula if you so wish it just may take you a little longer. I'm gonna go to the T. I. T. Four and if you go to stat and air over two tests and you go all the way down to the a option where it says one prop Z ent we're doing a one proportions E interval. And uh the X value remember was 2 75. That was the number of favorable the people that said that they follow college football out of a total possible 785 people that were randomly selected. And we want to be 95% confidence of 950.95 will be the sea level. And then whenever we calculate that this top band here that's our confidence interval. Now you can write the P. Hat if you want. It's about 35% but this top band is actually the answer. So I'll go in round I liked around three decimal places, so .317 or 31.7 And .384 or 38.4%. Now it doesn't actually say to do this. But if you wanted to interpret this, you would just say we can be Um, confident that the true population proportion of people who follow college football is between 31.7 and 38.4%.

We are very familiar with the methods. By now we want to construct the confidence level of 99 point Sorry off 99% and we have 20 different readings. So in order to control the confidence interval, what I've done is I've again simply input the values over here. These you can see we have 20 values. We are example mean a study 2.5 and a standard deviation is 20.3, you know, confidence level of 99. I select 99 this is the confidence interval that I get 20.8 to 44.2 20.8 20 pointed 20.8 to 44 point to This is my 99% confidence in trouble. Alright. Could also have done this by using the same old formula here and would have become 20 degree of freedom would be 19. Substitute these values Alfa. I would have become 0.50 point 00 fight and put the values in this formula and we come to the same result. This one now it is asked. Does this say? What is the question? There's the conference interval. Give us a good information about the population of all cancers off the same pretty brands that I consumed. No, because we have taken only one can off every brand in this, and only one can cannot be a representative off all the cans off that particular brand. So the answer is no. And we don't even care if this is normally distributed, because this is simply not going to give us a good idea off the, you know, off the mean off the mean off the mean of what we are. Yeah, the mean off caffeine for 12 rounds off a drink. So we don't even care if this is normally distributed.

Problem number 13. The same percentage deviation is the square root on some off the square, deviation from the need by one. So the deviation is equal to approximately opening to a 16 and the degrees of freedom, um, is equal toe end minus one, which is 17 minus one. So the chi square off one minus often Toto five. It is equal to 6.9 or eight and the car square off points also five is a 28.845 So the boundaries for the standard deviation is equal to end minus one off chi square off over two times s she is 17 minus one over 28 0.845 times 0, 00.2816 Mr. Approximately open toe one and the other boundary is in minus one over X square, one minus off over two times s which equal to 17 minus one over 6.9 or eight times open toe 816 which approximately opened 4 to 9. So the boundaries off the confidence interval for the variance is the square off his value, which is open toe one squared and opened 4 to 9 square, which is 0.44 and 4.184

Hello. This is a problem. 45 and first we need to do a one sample T test um to find the confidence interval for the population mean the average can be found through Excel And there are 20 numbers. Your sample size is 20 And the average that we got is 24.1. No, Once we do the one sample T test we need to input our sample mean In the sample start aviation which is 4.3 in our sample size. We write down that it's a confidence interval for form you And we give it a 90% confidence interval And we find that the 90% confidence interval is From 22.4 4- 25.76. If we want to find a 99% confidence into the only thing that changes is this part here And we get that the lower limit is 21.35 And the upper limit is 26.85. So the 90% confidence interval form you is going to equal 22.44 To 25.76. And for the 99 confidence interval whore. Um You It's going to be from 21.35 To 26.85. Yeah. And now let's interpret the confidence intervals with 90% confidence. In the meantime us adults spend watching television using DVRs each day is and interval 22.44 To 25.76. No. For the other confidence sensual, it's the same thing with except the percentage with 99 confidence. The means time U. S. Adults spend watching television using to yours each day. Yes. And uh interval 21.35 to 26.85. Now if we take a look at the widths the west of the 90% confidence interval is 3.36 which is smaller then 5.5 for the 99% confidence interval.


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