00:01
This is kind of a hodgepodge of a bunch of different questions.
00:08
A sales of a product in a weekday is normally distributed with a random sample of 16 sales, has a mean of 50, a sample variance of 225, and we're being asked to construct a 98 % confidence interval estimate of the population mean.
00:27
So we're going to take that 50 plus or minus 2 .6025 times the square root of 225 over 16.
00:37
And that's going to give us an interval of 40 .241 to 59 .759.
00:45
Now we're going to redo this problem, but this time we're going to do a 98 % confidence interval when the variance is 100.
00:55
So 50 plus or minus 2 .6025 times the square root of 100 over 16 will give us an interval of 43 .494 to 56 .506.
01:09
In this next problem, a survey of 500 college students who missed at least one meeting in a statistic course found that 66 % missed two or more classes.
01:25
Construct a 95 % confidence interval for the population proportion of students who missed one class who will miss more than two classes.
01:32
So a 95 % confidence interval in this one.
01:38
So we will take 0 .66 plus or minus 1 .96 times the square root of 0 .66 times 0 .34 over 500.
01:48
And that's going to give us an interval of 0 .618 to 0 .702.
01:53
For the next problem, if a manager wants to estimate the mean life of batteries within plus or minus 20 hours, so that would be our margin of error.
02:05
With 95 % confidence, what sample size is needed if the process standard deviation is 100 hours? so we're going to take 20 and set that equal to 1 .96 times 100 divided by the square root of our unknown sample size.
02:20
And then we're going to solve for n.
02:22
So n is going to be 1 .96 times 100 divided by 20, the quantity squared.
02:31
And that's going to give us a sample size of 96 .04, which we would round up to 97.
02:41
In the next problem, we have a sample, a mu, excuse me, of 375 and a standard deviation of 100.
02:52
A sample of 64 indicates that the sample mean is 350.
02:58
And they would like us to test at significance level of 0 .05 if there's evidence that the mean life is different from 375...