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anonymous
 4 years ago
Consider a binomial experiment with n trials for which np>5 and nq>5. What is the value of the continuity correction when n=25; when n=230? Round answer to 3 decimal places.
anonymous
 4 years ago
Consider a binomial experiment with n trials for which np>5 and nq>5. What is the value of the continuity correction when n=25; when n=230? Round answer to 3 decimal places.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As the value of n increases, does the continuity correction value increase, decrease, or stay the same?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why did you close it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know how to do it, but Zarkon, nikvist and KingGeorge are online, you can either repost it or tag them on this thread.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i had to so tht i could post another question, how do i tag them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using '@' Example: @Confucious

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Hero @Zarkon @KingGeorge

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know how to do this either, sorry

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Ask Satellite, Myininaya, JamesJ or amistre64

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Can you give me a rundown of what the continuity correction is?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0I've never even heard of continuity correction. What course is this?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Definitely some kind of statistics course.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3From what I've found on wikipedia, if you use a continuity correction of 0.500 you almost always get a good approximation. But I don't think this is what you want.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My txtbk says that the continuity correction is for the upper and lower endpoints of an interval, this question doesn't have any about an interval there, that's why I don't know what to do with it.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3I think I have it. According to the bottom of page 313 of http://books.google.com/books?id=kNutPR9P3hoC&pg=PA313&lpg=PA313&dq=how+to+find+the+continuity+correction&source=bl&ots=EioM54saDN&sig=1nJZqwxIBC2UHYbFHfWzp_qZcSU&hl=en&sa=X&ei=sHV7T_nmJKq22gXd07CIAw&ved=0CE8Q6AEwBg#v=onepage&q&f=false, the continuity correction should be\[{0.5 \over n}\]So for your numbers...\[{0.5 \over 25}\approx0.020\]\[{0.5 \over 230}\approx0.002\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3So if \(n\) increases, continuity correction decreases.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0...I'll try this. Thank you so much.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Look at this one please. Suppose we have a binomial distribution with n trials and probability of success p. The random variable r is the number of successes in the n trials, and the random variable representing the proportion of successes is = r/n. (a) n = 51; p = 0.45; Compute P(0.30 ≤ ≤ 0.45). (Round your answer to four decimal places.) (b) n = 49; p = 0.29; Compute the probability that will exceed 0.35. (Round your answer to four decimal places.)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have 0.006 </= phat </= 0.009 but that looks completely wrong.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Judging by what the previous question was, I would assume we need to find the continuity correction again which is \[{0.5 \over 51}\approx0.0098\]and add/subtract this to the interval \([0.3, 0.45]\). So it becomes \(0.2902 \leq \hat p \leq 0.4598\) Using a normal distribution.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Looking at the example in the book I linked to, we also need to calculate the standard deviation which would be \[\sigma_{\hat p} = \sqrt{pq \over n}=\sqrt{.45(.55) \over 51} \approx0.3518\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Still going from the example, we just need to find\[P\left( {.2902.45 \over .3518} \leq z \leq {.4598 .45 \over .3518} \right)\]Where \(z\) is that variable thing you use for the normal distribution.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Simplifying a bit, we get\[P\left( .4542 \leq z\leq .0279 \right)\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Which I get to be about \(P\approx 0.1863\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got a different answer for the standard deviation. It's 0.0697.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3How did you calculate the standard deviation?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3I'm going almost entirely by what that book I linked to is telling me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's the textbook I'm using but a newer edition that doesn't have some of this stuff and to find z it's (x minus the mean) over the standard deviation. We don't have the mean.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3According to the book\[\mu_{\hat p} = p=0.45\]Which is what I was using for the mean

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3But you are correct about the standard deviation. I was using \(n=2\) for some reason. I now have \[\sigma_{\hat p} \approx 0.0697\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3If we recalculate using that, we want\[P\left( {0.29020.45 \over 0.0697} \leq z\leq {0.4598 .45 \over 0.0697} \right)\]\[\approx P\left( 2.2927 \leq z \leq 0.1406 \right)\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Which finally gets us that \[P\approx 0.5449\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3And that was just for part a.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the z scores, they have to be rounded to 2 decimal places and then we find the probability from a table giving the areas of the standard normal distribution. From that I got 0.5447. How did you get 0.5449?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3I was using all 4 decimal places and my calculator. If you're using the table and 2 decimal places, go with 0.5447

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. We find the continuity correction again for part (b) right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But here's the thing. There's no interval so I don't know if I should add or subtract.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Use the interval \[P\left(.35\leq \hat p\leq 9\times 10^{99}\right)\] Or a similar upper bound.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Just make it so large that it doesn't matter what you do. At values that high, the sum of the probabilities greater than \(9\times 10^{99}\) is small that it might as well be 0.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Although you said you have to use a chart and not a calculator. That might complicate things.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't get that. How can that number be used. It looks like it's too big to be even written out.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3We can use it because the probability that \(z \geq 10^{99}\) is virtually \(0\). So really, we're dealing with 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so it can be \[P(0.35\le p\hat \le 0)\] ? [I don't know why it came out like that]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Let's just forget I said we could do that for the moment, and continue with the actual problem (we'll come back to it soon). That means that the continuity correction is \[{0.5\over49}\approx0.0102\]Also, \[\mu_{\hat p} =0.29\]\[\sigma_{\hat p} \approx 0.0648\]Feel free to check my work.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Thus, we need to find\[P\left({.3398.35 \over .0648} \leq z \right)\]or\[P\left( 0.1574 \leq z \right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where did you get .3398 and 0.1574 from??

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3I subtracted the continuity correction from the lower interval to get .3398 .3398=.350.0102

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3And \[{.3398.35 \over 0.648} \approx 0.1574\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Now let's get back to what we were talking about previously. How were you taught to find \[P(0.1574 \leq z)?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the text has a table in it that gives all the probabilities to the left of a z score from 3.49 to 3.49. Once we find the z score, we round it to 2 decimal places (because that's how the table has it) and we find it on the table and take the probability that's there for the answer.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Then you want to find the probability for \(z=0.16\) and subtract it from 1.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.3Sorry if I'm sounding angry or impatient, I'm really not :) I do need to leave for dinner now though. I'm getting an overall probability of about 0.5625. If you're very close to this, just assume you're correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok thanks a lot for your help
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