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A researcher wishes estimate the proportion adults who have high-speed Intemet access What size sample should be obtained she wishes the estimate be within 03 with ...

Question

A researcher wishes estimate the proportion adults who have high-speed Intemet access What size sample should be obtained she wishes the estimate be within 03 with 99%6 conlidence she uses Drevious estimate of 562 b) she does not use any prior estimates?a) n =(Round up to the nearest integer:)b) n-L (Round Up to the nearest integer:)

A researcher wishes estimate the proportion adults who have high-speed Intemet access What size sample should be obtained she wishes the estimate be within 03 with 99%6 conlidence she uses Drevious estimate of 562 b) she does not use any prior estimates? a) n = (Round up to the nearest integer:) b) n-L (Round Up to the nearest integer:)



Answers

High-Speed Internet Access A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 99\% confidence if (a) she uses a 2004 estimate of 0.44 obtained from a Harris poll? (b) she does not use any prior estimates?

Solving party of this problem so is given edge 0.3 So they're Alpha by two can be calculated as their 0.1 by two, which is equal to that 0.5 which is equal to two points. 575 Going ahead, as we all know in formula, which I am writing here and is equal to P multiple tickets and one minus B general far by two by e to the power to so just putting the value Which are I given in the problem so I can write 0.635 Multiplication one minus 10.6 35 Multiplication 2.575 by 0.3 Holy square. And on solving it I get the value Age 1708 which is our answer for party now going to calculate the value of part B. So we have then Alpha by two is equal to There are 0.1 by that's zero point 01 by two, which is equal to their 0.5 which is equal to 2.575 going ahead and just putting the value here so I can write the formula edge and is equal to 0.25 multiplication and gentle for buy to buy e hold to the power to so just putting the value 0.25 Multiplication. 2.575 by point 03 Holy square. And on simplifying it, I get the answer at 1842 So I hope you understand the solution.

So it's given that C is equal to 90% Or is your .9? He is 0.02 or 2%. One? Is your .275? P. Two? Yes. 0.231. So now we can use the formula for e sample size and we know that the question is asking us for to solve for the confidence level. See one minus alpha, which is your point. Yeah. No. Yeah. And determine Z of all for over two which is Z of 0.0 fun using the normal probability table and the appendix. So they should give us is he Score C is 1.645. No, We also know that end is equal to to see seemed over squared times sp and if you plug in all the values trade given up on We get 2,551. Now for confidence level, C equals one month. Alfa really can then find sample size using no prior estimates to one have see see the two. Yes, so and Just 1/2 times 1.645 over 0.02, Just 3,382.

So in part A for the confidence level, C. Is one minus alpha. Which is what's your point 95 to determine C. Of the two. Which is a sea of 0.025. Using the normal probability table in the appendix we can find that CFC Is equal to 1.96 now. No and is equal to CFC. This is when P one and P two R. No since over e squared times. Mhm. Was for the right to speak sp two. Unless P. Two. Yeah. And just plugging in all of these values that we already know. 96 over. Sure. Cancer three Time 0.2. Yeah. Yeah. Two you're and seven which gives us approximately 1406. Now for the confidence level C. As one man sequel to to open 95 and see alpha two physical says you're take the dizzy what's your point 0 to 5? We can use the normal probability Table in the Appendix to get c. 1.96 and therefore an is equal to two times you see. Mhm. Spread because he puts 1/2 times one thing 96 over your country. Three Spring, which is a good too, 2000 100 good for.

Experiments to make sure that we have the right standard error. You have a researcher who wants to give a 95% confidence interval with an error of about 4% for the number of people who go abroad each year. Now we know that last year 40 people out of 200 surveyed went abroad, and we want to know how many people. So we survey so that we can meet these requirements well, before we had that are standard error is equal to R Z score times the square root over estimate for P R estimate for Q over end. And if we do a little bit of rearranging, we get that end is equal to rs mitt for P hat s me for Q hat times R Z score over the standard error squared and this the formula we're gonna be using for this problem. What's P Hat P hat were given 40 over 200 and cue hat is gonna be 200 minus 40 all over 200 great r Z score. Since we're doing a 95% confidence interval is going to be 1.96 and our standard error is gonna be 0.4 for that 4%. We take all of that and plug it into a calculator. We get 384 0.16 and we're gonna round that right on up to 385 for the number of people we need to ensure the specific confidence interval that we wants. Now, this is a nice, ideal case. But what if we don't know how many people went abroad last year, which might really be the truth. Well, instead of assuming that it's 40 at 200 we're gonna assume that it's 100 out of 200 even split, So we're gonna take the same formula. But instead of P hat being 40 over 200 now it's 100 over 200. 1/2 que hat is gonna be the same 100 over 200 and everything that's getting squared is going to stay the same 1.96 over 0.4 squared. So the only things that have changed r p hat and Q hat values and if we plug that into a calculator, we got 600.25 much bigger than it was before, because when we don't have any information, we need to ask more people just to be sure. And we're going around this up again, can't have 1/4 of a person. So if we don't know anything, we have to interview 601 people in our study.


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