1. ## confidence intervals

question:the claimed germination rate of seeds in a packet is 90%. a gardener suspects the real rate is not as high as it is claimed so he carries out a trial. he selects 120 seeds and 97 germinate. test to see if his suspicions are well founded by using a confidence interval for the population portion at the 5% significance level.

so what i would of thought the right answer would be is saying the population portion=97/120 and then using the formula margin of error formula 1/√120 la . +/- this to the population portion but apparently this is wrong. apparently you have to use the formula √[(p(1-p))/n] i have no idea where this formula comes from and why you have to use it. I was only thought about the 1/√n formula so 1 what is this formula? 2 why do you use it here? and 3 how can I tell which formula to use ?

thanks

2. ## Re: confidence intervals

Check out the wikipedia article:
https://en.wikipedia.org/wiki/Binomial_distribution
Specifically the section for Confidence Intervals and the Wald Method. You are not using p hat, you are using p. You want to see if your value of 97/120 is within two standard deviations of the given germination rate. So, you have:

$.9 \pm 2\sqrt{\dfrac{.9\cdot .1}{120}}$

which gives the approximate interval: $.8452277 \le \hat{p} \le .954772$.

The observed test statistic of $\dfrac{97}{120} \approx .808333$ is not within that range, leading you to conclude that you are 95% confident that the claimed germination rate is false.

Now, replacing p with p hat (using your test statistic), you can create a new confidence interval. You are 95% confident the actual germination rate is between .73647 and .88020.

3. ## Re: confidence intervals

sorry I forgot to mention that √[(p(1-p))/n] was multiplied by 1.96 and I am not sure where this comes from as well.

i only have a basic statistics understanding. i tried to read that wiki stuff but it just washed over my head as they use terms I do not know of to explain other terms. I am reading the section you told me to read and I am having trouble understanding what z is (what the probit is)

do you know how I can distinguish when to use z √[(p(1-p))/n] and when to use 1/√n because in the questions i am doing they are using these on different occasions

thanks

4. ## Re: confidence intervals

Originally Posted by edwardkiely
sorry I forgot to mention that √[(p(1-p))/n] was multiplied by 1.96 and I am not sure where this comes from as well.

i only have a basic statistics understanding. i tried to read that wiki stuff but it just washed over my head as they use terms I do not know of to explain other terms. I am reading the section you told me to read and I am having trouble understanding what z is (what the probit is)

do you know how I can distinguish when to use z √[(p(1-p))/n] and when to use 1/√n because in the questions i am doing they are using these on different occasions

thanks
For confidence intervals, you will always use standard deviation. -1.96 is the z-score corresponding to 2.5% of the bell curve. 1.96 is the z-score corresponding to 97.5% of the bell curve. So, if you take an interval between them, you are covering 97.5%-2.5% = 95% of the bell curve. That leaves 5% unaccounted for. That is where the term 5% significance level. There is a 5% chance that you are wrong. Suppose you wanted a 17% significance level. Then you would want half of that at the left side and half of that at the right side of the bell curve. So, you would be looking for the z-score associated with 8.5% of the bell curve. That is a z-score of -1.37. Then you want the z-score associated with 100%-8.5%=91.5%, which is a z-score of 1.37 (I just used a chart to find the correct z-score values).

Edit: It is fairly common practice in introductory statistics courses to use "about" plus or minus 2 standard deviations as the 95% confidence interval range (-1.96 is very close to -2 and 1.96 is very close to 2).

5. ## Re: confidence intervals

thanks I understand now where the 1.96 comes from but I still do not know how I can distinguish when to use z √[(p(1-p))/n] and when to use 1/√n b

6. ## Re: confidence intervals

Originally Posted by edwardkiely
thanks I understand now where the 1.96 comes from but I still do not know how I can distinguish when to use z √[(p(1-p))/n] and when to use 1/√n b
When you are working with confidence intervals, you use standard deviation: √[(p(1-p))/n]. When you are working with standard error, you use standard error: 1/√n.

7. ## Re: confidence intervals

Originally Posted by SlipEternal
When you are working with confidence intervals, you use standard deviation: √[(p(1-p))/n]. When you are working with standard error, you use standard error: 1/√n.
I wish I could be more specific, but for a beginner statistics course, you will not have the necessary mathematical knowledge to know when else you might use one or the other. For now, it will be easiest to just memorize. When you get to more advanced mathematics, you can learn more advanced techniques for figuring out the correct formula.

8. ## Re: confidence intervals

I understand it better now.

thanks