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

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- schrodinger

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- anonymous

As the value of n increases, does the continuity correction value increase, decrease, or stay the same?

- anonymous

Why did you close it?

- anonymous

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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## More answers

- anonymous

i had to so tht i could post another question, how do i tag them?

- anonymous

Using '@'
Example: @Confucious

- anonymous

thanks

- anonymous

@Hero
@Zarkon
@KingGeorge

- Hero

I don't know how to do this either, sorry

- Hero

Ask Satellite, Myininaya, JamesJ or amistre64

- anonymous

k thanks

- KingGeorge

Can you give me a rundown of what the continuity correction is?

- Hero

I've never even heard of continuity correction. What course is this?

- KingGeorge

Definitely some kind of statistics course.

- KingGeorge

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.

- anonymous

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.

- KingGeorge

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...\[{0.5 \over 25}\approx0.020\]\[{0.5 \over 230}\approx0.002\]

- KingGeorge

So if \(n\) increases, continuity correction decreases.

- anonymous

...I'll try this. Thank you so much.

- KingGeorge

You're welcome.

- anonymous

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 â‰¤ â‰¤ 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

I have 0.006

- KingGeorge

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.

- KingGeorge

Looking at the example in the book I linked to, we also need to calculate the standard deviation which would be \[\sigma_{\hat p} = \sqrt{pq \over n}=\sqrt{.45(.55) \over 51} \approx0.3518\]

- KingGeorge

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.

- KingGeorge

Simplifying a bit, we get\[P\left( -.4542 \leq z\leq .0279 \right)\]

- KingGeorge

Which I get to be about \(P\approx 0.1863\)

- anonymous

I got a different answer for the standard deviation. It's 0.0697.

- KingGeorge

How did you calculate the standard deviation?

- KingGeorge

I'm going almost entirely by what that book I linked to is telling me.

- anonymous

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.

- KingGeorge

According to the book\[\mu_{\hat p} = p=0.45\]Which is what I was using for the mean

- KingGeorge

But you are correct about the standard deviation. I was using \(n=2\) for some reason. I now have \[\sigma_{\hat p} \approx 0.0697\]

- KingGeorge

If we recalculate using that, we want\[P\left( {0.2902-0.45 \over 0.0697} \leq z\leq {0.4598 -.45 \over 0.0697} \right)\]\[\approx P\left( -2.2927 \leq z \leq 0.1406 \right)\]

- anonymous

oh yh

- KingGeorge

Which finally gets us that \[P\approx 0.5449\]

- KingGeorge

And that was just for part a.

- anonymous

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?

- KingGeorge

I was using all 4 decimal places and my calculator. If you're using the table and 2 decimal places, go with 0.5447

- anonymous

Okay. We find the continuity correction again for part (b) right?

- KingGeorge

yup.

- anonymous

But here's the thing. There's no interval so I don't know if I should add or subtract.

- KingGeorge

Use the interval \[P\left(.35\leq \hat p\leq 9\times 10^{99}\right)\] Or a similar upper bound.

- anonymous

??? why???

- KingGeorge

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.

- KingGeorge

Although you said you have to use a chart and not a calculator. That might complicate things.

- anonymous

I don't get that. How can that number be used. It looks like it's too big to be even written out.

- KingGeorge

We can use it because the probability that \(z \geq 10^{99}\) is virtually \(0\). So really, we're dealing with 0.

- anonymous

so it can be \[P(0.35\le p-\hat \le 0)\] ?
[I don't know why it came out like that]

- KingGeorge

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.

- KingGeorge

Thus, we need to find\[P\left({.3398-.35 \over .0648} \leq z \right)\]or\[P\left( -0.1574 \leq z \right)\]

- anonymous

where did you get .3398 and -0.1574 from??

- KingGeorge

I subtracted the continuity correction from the lower interval to get .3398
.3398=.35-0.0102

- KingGeorge

And \[{.3398-.35 \over 0.648} \approx -0.1574\]

- KingGeorge

Now let's get back to what we were talking about previously. How were you taught to find \[P(-0.1574 \leq z)?\]

- anonymous

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.

- KingGeorge

Then you want to find the probability for \(z=-0.16\) and subtract it from 1.

- KingGeorge

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.

- anonymous

ok thanks a lot for your help

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