Question

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.

Answer 1

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

Answer 2

Why did you close it?

Answer 3

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.

Answer 4

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

Answer 5

Using '@'
Example: @Confucious

Answer 6

thanks

Answer 7

@Hero
@Zarkon
@KingGeorge

Answer 8

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

Answer 9

Ask Satellite, Myininaya, JamesJ or amistre64

Answer 10

k thanks

Answer 11

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

Answer 12

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

Answer 13

Definitely some kind of statistics course.

Answer 14

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.

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

You're welcome.

Answer 20

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.)

Answer 21

I have 0.006 </= p-hat </= 0.009 but that looks completely wrong.

Answer 22

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.

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

How did you calculate the standard deviation?

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

oh yh

Answer 35

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

Answer 36

And that was just for part a.

Answer 37

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?

Answer 38

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

Answer 39

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

Answer 40

yup.

Answer 41

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

Answer 42

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

Answer 43

??? why???

Answer 44

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.

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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.

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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.

Answer 56

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

Answer 57

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.

Answer 58

ok thanks a lot for your help