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Do Homcuton Irancslo Dancol quogle Chromemathxlcom StudenuplayerHomeirorkaspxchomeworkId tel;e omestonide3oaj _5Math 227. Fall 2019,HO7 , Section 18697Homework: Sec...

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Do Homcuton Irancslo Dancol quogle Chromemathxlcom StudenuplayerHomeirorkaspxchomeworkId tel;e omestonide3oaj _5Math 227. Fall 2019,HO7 , Section 18697Homework: Sec. 6.3 tati Score: 4 of 8 complete)SaveHW Score: 38.89%,6.3.17-TQuestion HelpAtesearcher Mshes estimate, with 9916 confidence, the population praporiion of adults who think Congress doing Dood cecclleni Job. Her eslimate MuSL be accurato within 590 01 Ihe true proportion; (a) No preliminary estirale available. Find Ihe minimum sample s

Do Homcuton Irancslo Dancol quogle Chrome mathxlcom StudenuplayerHomeirorkaspxchomeworkId tel;e omestonide 3oaj _5 Math 227. Fall 2019,HO7 , Section 18697 Homework: Sec. 6.3 tati Score: 4 of 8 complete) Save HW Score: 38.89%, 6.3.17-T Question Help Atesearcher Mshes estimate, with 9916 confidence, the population praporiion of adults who think Congress doing Dood cecclleni Job. Her eslimate MuSL be accurato within 590 01 Ihe true proportion; (a) No preliminary estirale available. Find Ihe minimum sample size needod: (b) Find the minimum sample sie needed; using prior study Inal (ound Inal 4690 of the respondents sard Iney thunk Congress doine good oycelleni Compare Ine resulls Iromn parts (a) and (0) Vour (a) Wnal / Ihe minimum sample size needed assuming that no prior inforalion available? n=L(Round up Ine ncarest Who @ number neoded,) Enter your answer in the answer box and then click Check Answer: parts remaining



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For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Heights of 1-Year-olds The average 1 -year-old (both genders) is 29 inches tall. A random sample of 301 -year-olds in a large day care franchise resulted in the following heights. At $\alpha=0.05,$ can it be concluded that the average height differs from 29 inches? Assume $\sigma=2.61$ $$ \begin{array}{lllllllllll}{25} & {32} & {35} & {25} & {30} & {26.5} & {26} & {25.5} & {29.5} & {32} \\ {30} & {28.5} & {30} & {32} & {28} & {31.5} & {29} & {29.5} & {30} & {34} \\ {29} & {32} & {27} & {28} & {33} & {28} & {27} & {32} & {29} & {29.5}\end{array} $$

The following is a solution to number five and this deals with proportions. And the minimum sample size is necessary in order to satisfy certain conditions. So on this first one, we're asked to find to find the minimum sample size necessary to be 80% confident with a margin of error, no more than 5% points. And there are two parts one where we know The P one and P two and the other one where we don't know. So whatever, we don't know, we have to use .5 because that maximizes are in. So the less, you know, the larger end must be, so in equals the Z star, which is the critical value square and that's 1.282. So uh we square that. Now if you don't know how to find that that you can get in a table but you can also get in the calculator. Now, normally I say go ahead and memorize this, these Z scores, but 80% is kind of an uncommon one. I mean it's, You should know it, but just in case you just make the area .8, The mean is zero. The standard deviations one and just make sure it's center, so you want this center here and then you paste and that's where I get that 1.282. Okay, so that's that part. And then in here. So since I don't know what P one and P two are, I'm just gonna save 20.5 and 0.5 for p one and one minus P one and then plus P two is also 20.5 and then one minus 10.5 is also 0.5. And then I divide that by the margin of error square 2.5 squared. And whenever you plug that in you should get Uh once I find it here so 328.7. But then you always always always round up to the next hole number because we can't have in decimals as sample sizes. So it should be 329. Okay the next part is where you actually know what P. One and P two R. And we set it up the same way but we're actually gonna use the values so 1.282 is still disease score. And then here we're gonna say 0.2 times 0.8 so one minus 10.2 is 0.8. And then plus this P two is 20.65 and then one minus 10.65 point +35. And then we divide that By the margin of error which remember was .05 squared. Okay so whenever you plug that in you should get 254 .7 which of course rounds up to 255. So no decimals whenever you're looking for in. Okay. Part two is the 90% confidence level with a margin of error of 0.2. So again P one P two are unknown. So I'm gonna use 20.5 and equals 1.645 is the Z score for 90% And then .5 times five Plus five times 5. Since I don't know anything. And then divide that by .2 squared. And he plugged that in the calculator should get quite a large number here. So 33 82 0.5. And then of course round that up to 33 83. That's the minimum sample size necessary. Mm. Okay. And then let's say you do know what P one and P two are. Well, same thing. We're gonna do 1.645. Since we're still wanting to be 90% confident. And then this time we're actually going to use those numbers. So .75 and then 1 -175 which would be .25 and then plus six three and then one minus +63 is +37. And then we divide that By that standard air of .2 squared. Okay, so you plug that in And should get 28 45.3. Now, we're always talked around down in the you know, in early math button statistics, we don't do that. We always round up. So it's gonna be 28 46. Always round up. Never round off. Even if it's less than than the 5:28 46. Okay. And then the last 1 95% confident. So that means that Z score is 1.96. We're gonna square that. And since we don't know what P one and P two R. I'm gonna use 20.5 point five .5.5. And then we divide that by the air squared which is .10 squared. And then whenever you plug that in you should get 192.0 Zero, something like maybe 02 or something like that. You still round up. I know it seems you know way off. We're taught early on. Okay, well that's just 192. But that's not the case. We round up up up if there's a decimal and also 92.02 goes up to 193. And then the next part is where we actually know what the the P one and P two R. So we still use the 1.96 squared since we still want to be 95% confidence And we're gonna save .11 times .89 Plus 3 7 times 6 3. And then we divide that by that standard air of 0.10 square or the margin of air. And that gives us 127 .16, which of course goes up to 1 28. So these were the necessary, the minimum necessary uh sample sizes. If we wanted to satisfy those following conditions

The following is a solution to number eight. And we're looking at the difference of mean G. P. A among all men in a college affiliated with frats or not affiliated or unaffiliated from the fraternities and we want to be 95% confidence confident with a margin of error of 950.15. And fortunately the standard deviations for the two populations are the same .4. So we're gonna use this formula, we're in. Now let me change colors like blue. There we go. Okay, so in is the critical value of 95% confidence, which would be 1.96 and we're gonna square that. So if you don't have those Z scores memorize you really should, you can get it from a table or calculator. And we're gonna multiply that by 0.4 squared plus 0.4 squared. So the two variances added together, which would be 20.4 squared. Then we divide that by the the error hopes .15 and we square that. And when we do that, we should get 54.6. Now you always always, always, even if it's less than five, you always round up to the nearest whole number because your sample size needs to be a whole number. So we were we will round this up to 55. So 55 is the minimum sample size necessary to satisfy these conditions

Hi. We're gonna start with stating our null hypothesis, which is at the mail. Old mule is equal to, uh, $5400 on our claim, which is the alternative. Is that that the mean is Liston $5400 on the critical value. We can find it from the table. Ah, As, um minus two point or 52 on. That's why. Because the degree of freedom is 27 on. We are looking at a point or 25 and as you know, it is one Tim. And it just looked it and that this value is Ah, this one. And we can commuted by this for my expose. What is that? Yeah, the average that we find and minus a mule divided by s, which is assembled. Send a deviation, divide by square, root off, and you find out that it is minus 1.262 on. Here's a graphical AIDS toe to visualize the idea. So here's the critical value and anything be lowered. It's gonna be, um, and the rejection area. But we happens to be greater than that. The the critical value. So we don't reject the now on and for summarizing the results, You can say there is no enough evidence to support the claim that the average course is the Stan 54 $5400.

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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