Suppose the following data are selected randomly from a population of normally distributed values.
40
51
43
48
44
57
54
39
42
48
45
39
43
Construct a 95% confidence interval to estimate the population mean.
n =13. X bar = 45.62
5 = 5.694
Well, I'm not usually one for serious numbers, but I'll give it a shot! Based on the data provided, we can calculate the standard deviation (s) using the formula:
s = √(Σ(x - X bar)^2 / (n - 1))
Using the given values, we can compute the standard deviation to be approximately 5.694.
Now, to construct a 95% confidence interval for the population mean (μ), we can use the formula:
CI = X bar ± (t * (s/√n)
Where X bar is the sample mean, t is the critical t-value for a 95% confidence interval with n-1 degrees of freedom, s is the standard deviation, and n is the sample size.
Since the sample size is 13, we have n-1 = 12 degrees of freedom. The critical t-value for a 95% confidence interval with 12 degrees of freedom is approximately 2.201.
Plugging in the values, we get:
CI = 45.62 ± (2.201 * (5.694/√13))
Now all we have to do is calculate!
After some long calculations, the 95% confidence interval for the population mean is approximately 41.22 to 49.02.
So, we can be 95% confident that the population mean falls between 41.22 and 49.02. Keep in mind, though, that this whole statistical talk may not make much sense to the clown inside of me!