Question 1176562: A random sample of 8 observations was drawn from a normal population. The sample mean and
sample standard deviation are X = 40 and s = 10.
a) Estimate the population mean with 95% confidence.
b) Repeat part (a) assuming that you know that the population standard deviation is σ = 10.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
The sample size is n = 8 and we don't know sigma, which is the population standard deviation. Instead, we have the sample standard deviation value s = 10. This sample statistic estimates the population parameter.
Because n > 30 is not true, and we don't know sigma, we must use the T distribution.
We have n-1 = 8-1 = 7 degrees of freedom.
Use a T table such as this one
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
At the bottom of the table it shows the various confidence levels. Locate the 95% confidence level column.
Then mark the df = 7 row
This is what you should have
We can see the value 2.365 is at the intersection of the row and column we highlighted.
The t critical value is roughly t = 2.365
Let's compute the lower bound L
L = xbar - t*s/sqrt(n)
L = 40 - 2.365*10/sqrt(8)
L = 40 - 8.362
L = 31.638
L = 31.64
Now the upper bound U
U = xbar + t*s/sqrt(n)
U = 40 + 2.365*10/sqrt(8)
U = 40 + 8.362
U = 48.362
U = 48.36
The 95% confidence interval in the form (L, U) is (31.64, 48.36)
Answer: (31.64, 48.36)
We can write this in the form L < mu < U to say 31.64 < mu < 48.36
This format is more descriptive in that it's more clear that we're estimating mu here.
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Part (b)
Now we're told that sigma = 10, while everything else is kept the same.
Since we know sigma, we can use the Z distribution this time.
Using a Z table, the critical value is roughly z = 1.960 at 95% confidence.
Lower Bound
L = xbar - z*sigma/sqrt(n)
L = 40 - 1.960*10/sqrt(8)
L = 40 - 6.930
L = 33.07
Upper Bound
U = xbar + z*sigma/sqrt(n)
U = 40 + 1.960*10/sqrt(8)
U = 40 + 6.930
U = 46.93
As you can see, the format and structure of each formula is pretty much identical to the T distribution variety used in part (a). The only difference is that t has been replaced with z (so 2.365 is replaced with 1.960), and that we used sigma in place of s.
Answer: (33.07, 46.93)
This is equivalent to saying 33.07 < mu < 46.93
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