Confidence intervals, proportions

#1
A simple random sample of 1000 New Yorkers finds that 87 are left-handed.
(a) Find the 95% confidence interval for the proportion of New Yorkers who are left-handed.
The answer my professor gave and the answer that I got are completely off from each other.
Here is what I have so far
n = 1000
p^ = 87/ 1000 = 0.087
p^ ± z * Sqr. rt p^(1 − p^ ) / n

= 0.087 ± 1.960 ∗ sqr.rt 0.087(1 − 0.087)/1000

0.087 + 1.960 ∗ sqr.rt 0.087(1 − 0.087)/1000= .0089124071
0.087 - 1.960 ∗ sqr.rt 0.087(1 − 0.087)/1000= .01824369730

I looked up videos and they show that one is supposed to stop here , where the interval is just between the calculated numbers.

This is the correct answer that my professor gave:
.087 ± 0.0175 = (0.0695,0.1045)

I don't know how to get to 0.0175. I tried to see if its connected to the z chart somehow but I'm stuck.
Can anyone help me understand this? Or let me know if I'm doing anything wrong? :confused:
Thanks :wave:
 
Last edited:

Dason

Ambassador to the humans
#2
Looks like you might just be doing the algebra wrong?

Using the R programming language...
Code:
> p <- 87/1000
> 1.96 * sqrt(p*(1-p)/1000)
[1] 0.01746832
That last one looks like what you should be getting