00:01
In this exercise, we are provided with the data for a sample of size 6.
00:07
And for part a, we were asked to first check the assumption that the population is normally distributed.
00:15
So we can run a check based on the sample data.
00:19
And we can do this by constructing a normal probability plot.
00:24
I've done this in r, so i will show you the steps that i used in r to do this.
00:29
So the first step is to enter the sample data.
00:31
So i put it in a variable called x.
00:34
So all of this data here is the data provided in the question.
00:39
And then i use the function qq norm and the argument x.
00:45
And that makes a normal probability plot of the sample data.
00:51
If the sample data is normally distributed, then you see a fairly straight line from the bottom left to the top right, which is the case here.
00:59
So the assumption of normality of the underlying population is reasonable.
01:28
Now, here when i talk about the underlying population, for the specific question, it is the polyunsaturated fatty acid levels in the margin of a particular brand, the diet margin of a particular brand.
01:44
So the underlying population is all the margin that that brand makes.
01:51
For b, we are asked to find a 90 % confidence interval for the population mean and provide an interpretation of this interval.
02:02
Now, if we don't know the standard deviation of the population, which is the case here, a confidence interval for the population mean can be given by this formula, and we can calculate most of these parameters from the sample data.
02:21
So first, the sample average.
02:24
This can be found simply by using the mean function and using x as the argument.
02:30
It's 16 .983.
02:40
The sample size is 6.
02:44
And the sample of standard deviation, it can be found by using the function sd and the argument x.
02:54
We get 0 .3189.
03:03
The only thing left is the relevant percentile...