00:01
We have three problems to talk about.
00:03
How many women must be randomly selected to estimate the mean weight of women in one age group? we want 90 % confidence that the sample mean is within 3 .2 of the population mean, and the population standard deviation is known to be 16 pounds.
00:24
So within 3 .2, which we would call our error, and the standard deviation is 16 pounds.
00:33
So we're going to use this formula and equals our z value times our standard deviation divided by our error.
00:58
So when i'm dealing with 90 % confidence, that's value is 1 .64 times my standard deviation, which is 16, divided by my error, which is 3 .2.
01:13
And then i'm going to square this.
01:18
So i'm going to square this.
01:20
So n equals 8 .225 squared.
01:24
So the number is 67 .65.
01:29
That's, i would need to randomly select about 68 women for that calculation.
01:39
So number two, in a random sample of 170 college students, so n equals 170, 120 had part -time jobs.
01:51
Find the margin of error for the 95 % confidence interval used to estimate the population proportion.
02:01
Let's start by finding p -hat.
02:04
P -hat is 112 divided by 170, which is 0 .6588.
02:10
So now we can calculate our z value by taking p -hat times 1 -1 -p -hat, dividing it by our sample size and then taking the square root.
02:27
So z equals 0 .3064, which would be 3 .64 percent...