00:02
In this problem, you have a sample of an equals 1 ,000, and it's obtained from a population of a million.
00:12
Your p is 0 .35 for your population.
00:17
It says, describe the sampling distribution of p hat.
00:21
And so i wrote down the information that i just read over.
00:27
So your p is 0 .35, and you can estimate p hat to be 0 .35 also.
00:35
What you want to do is you want to check and see if n times p and n times k are both greater than five.
00:42
And if so, then you'll be able to say that it will look like a normal distribution.
00:47
And so if you do a thousand times 0 .35, we get 350.
00:57
And then if you do 1 ,000 times 0 .35, you get 0 .6.
01:01
I mean, times 0 .65, you get 650.
01:03
So both np and nq are both greater than 5.
01:07
And therefore, you can say the p -hat distribution is approximately normal, and your view for the p -hat would equal p -and your standard deviation for your p -hat distribution would equal p -q divided by n.
01:40
Then part b says what is the probability of obtaining x equals 390 or more individuals with the characteristic? well, since we are dealing with proportions in this problem instead of whole numbers, then we need to find out what proportion that is talking about.
02:01
And so 390, we would take and divide that by 1 ,000 to get the proportion.
02:06
And do that, you get 0 .39.
02:26
And so the question was asking, was it probably maintaining 390 or more? so we want then p -hat to be greater than or equal to 0 .39.
02:52
And when we start with p -hat interval, in order to use the normal curve to approximate this value, we need to then change this to an x interval.
03:01
We want to change that to the probability that x is greater than or equal to .39.
03:09
I mean, sorry, not .39.
03:13
We need to find out the value that is going to go in there.
03:18
And to do that, we need to apply a continuity correction.
03:24
The continuity correction for p hat is .5 over n.
03:34
And in this case, that would be .5 divided by 1 ,000.
03:42
Which would give us 8705.
03:57
And to apply the continuity correction, what we need to do is add that to high numbers and subtract it from low numbers.
04:07
Well, if you think about 0 .39 on the number line, there's 0 .0s down here.
04:18
Here's 0 .39.
04:20
We're talking about greater than equal to 0 .39.
04:23
So we're wanting here in this way.
04:27
And so that makes 0 .39 a low number.
04:30
And so we need to subtract it.
04:34
And so we'll do 0 .39 minus 0 .005.
04:50
And you're going to get the value for x that you're going to use, which is 0 .3895.
05:08
After that is you are going to change that to a z interval so that you can look it up on the standard normal table...