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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.

Mathematics
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As the value of n increases, does the continuity correction value increase, decrease, or stay the same?
Why did you close it?
I 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.

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i had to so tht i could post another question, how do i tag them?
Using '@' Example: @Confucious
thanks
I don't know how to do this either, sorry
Ask Satellite, Myininaya, JamesJ or amistre64
k thanks
Can you give me a rundown of what the continuity correction is?
I've never even heard of continuity correction. What course is this?
Definitely some kind of statistics course.
From 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.
My 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.
I 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...\[{0.5 \over 25}\approx0.020\]\[{0.5 \over 230}\approx0.002\]
So if \(n\) increases, continuity correction decreases.
...I'll try this. Thank you so much.
You're welcome.
Look 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.)
I have 0.006
Judging 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.
Looking at the example in the book I linked to, we also need to calculate the standard deviation which would be \[\sigma_{\hat p} = \sqrt{pq \over n}=\sqrt{.45(.55) \over 51} \approx0.3518\]
Still 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.
Simplifying a bit, we get\[P\left( -.4542 \leq z\leq .0279 \right)\]
Which I get to be about \(P\approx 0.1863\)
I got a different answer for the standard deviation. It's 0.0697.
How did you calculate the standard deviation?
I'm going almost entirely by what that book I linked to is telling me.
That'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.
According to the book\[\mu_{\hat p} = p=0.45\]Which is what I was using for the mean
But you are correct about the standard deviation. I was using \(n=2\) for some reason. I now have \[\sigma_{\hat p} \approx 0.0697\]
If we recalculate using that, we want\[P\left( {0.2902-0.45 \over 0.0697} \leq z\leq {0.4598 -.45 \over 0.0697} \right)\]\[\approx P\left( -2.2927 \leq z \leq 0.1406 \right)\]
oh yh
Which finally gets us that \[P\approx 0.5449\]
And that was just for part a.
For 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?
I was using all 4 decimal places and my calculator. If you're using the table and 2 decimal places, go with 0.5447
Okay. We find the continuity correction again for part (b) right?
yup.
But here's the thing. There's no interval so I don't know if I should add or subtract.
Use the interval \[P\left(.35\leq \hat p\leq 9\times 10^{99}\right)\] Or a similar upper bound.
??? why???
Just 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.
Although you said you have to use a chart and not a calculator. That might complicate things.
I don't get that. How can that number be used. It looks like it's too big to be even written out.
We can use it because the probability that \(z \geq 10^{99}\) is virtually \(0\). So really, we're dealing with 0.
so it can be \[P(0.35\le p-\hat \le 0)\] ? [I don't know why it came out like that]
Let'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.
Thus, we need to find\[P\left({.3398-.35 \over .0648} \leq z \right)\]or\[P\left( -0.1574 \leq z \right)\]
where did you get .3398 and -0.1574 from??
I subtracted the continuity correction from the lower interval to get .3398 .3398=.35-0.0102
And \[{.3398-.35 \over 0.648} \approx -0.1574\]
Now let's get back to what we were talking about previously. How were you taught to find \[P(-0.1574 \leq z)?\]
the 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.
Then you want to find the probability for \(z=-0.16\) and subtract it from 1.
Sorry 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.
ok thanks a lot for your help

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