00:01
On this scenario, we're going to agree that 35 % of people are going to be padding their insurance claim.
00:06
And we want to find, we have a sample of 138, and we want to know what is the likelihood that more than 50 % of them are in that category of padding.
00:17
So we can use our sampling distribution of p hat, and we can say that our proportion is going to be centered.
00:25
So the mean for p -hats would be at 0 .35, and then our standard deviation for p -hats would end up being 0 .35 times 0 .65 divided by the sample size of 138.
00:40
And we can do that and find what the likelihood of being higher than 0 .5 is.
00:45
So this is one technique, probably the most common technique.
00:48
We alternately could use binomial probability.
00:51
So if you're on a binomial probability section, we probably want to change this in 20.
00:56
Count and deal with binomial probability.
00:58
But i'm going to assume you're using the sampling distribution.
01:02
So we convert it to a z value.
01:04
We have 0 .5 minus 0 .35 divided by, and again, that standard error of 0 .35 times 0 .65 divided by that 138.
01:17
And so we have 0 .15 divided by square root of 0 .35 times .65 divided by 138.
01:27
And that gives us a test statistic, a z value of 3 .69.
01:35
And the area above 3 .69 is the same as that below negative 3 .6, negative 3 .69.
01:44
And that's approximately zero.
01:46
It's very small.
01:47
Doesn't even show up on my table.
01:49
So it's very unlikely.
01:50
Now on part b, we want to define what's the probability that the number of claims would be less than 45.
02:00
And the number of claims less than 45 is having a p hat that is less than 45 over 138.
02:09
And so all i need do is replace this here.
02:14
And so i'm going to go back on my and convert this now to a z.
02:20
Value...