00:01
For this question, we are given the data for a sample of four bags of carrots, which were all labeled 20 pounds, and we're told that the distribution of weights is normal, and based on the sample we are asked to find a 95 % confidence interval for the mean weight of all such bags of carrots.
00:20
Now since the population is normally distributed, and since we don't have the population standard deviation, we're going to rely on calculating the sample standard deviation, the confidence interval for the population mean is given as the sample mean plus or minus a critical value t sub alpha over 2 with n minus 1 degrees of freedom times the sample standard deviation over the square root of the sample size.
00:49
Since we want 95 % confidence, this means that alpha is 1 minus 0 .95 which is 0 .05 and so our critical value is t sub 0 .025 with n minus 1 or 3 degrees of freedom.
01:07
And using the software this is found to be approximately 3 .182.
01:16
And so now let's calculate the sample mean and sample standard deviation.
01:20
To do this, i've gone ahead and put the sample data in excel.
01:24
So in column a, we have the four data points for the sample.
01:28
To calculate the mean of the sample, start a computation with equals.
01:33
The function we want to use is the average function.
01:36
And for the arguments, we select the data points.
01:37
Hit enter.
01:39
And we get an average of 20 .275.
01:42
For the sample standard deviation, the function is std -ev .s.
01:48
That's the function we want...