Question 1207748: Construct the indicated confidence interval for the population mean mu using the t-distribution. Assume the population is normally distributed.
cequals0.98, x overbarequals13.6, sequals0.94, nequals19
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
xbar = sample mean
s = sample standard deviation
n = sample size
df = degrees of freedom
xbar = 13.6
s = 0.94
n = 19
df = n-1 = 19-1 = 18
Use a T table such as this one
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Similar tables should be found in the back of your stats textbook.
Use that table to determine the t critical value is roughly t = 2.552 at a 98% confidence level and when df = 18.
The confidence levels are labeled at the bottom.
What this indicates is that P(-2.552 < T < 2.552) = 0.98 approximately when df = 18.
Let's compute the margin of error.
E = t*s/sqrt(n)
E = 2.552*0.94/sqrt(19)
E = 0.55034082 approximately
Use this to find the lower and upper bounds (L and U)
L = xbar - E
L = 13.6 - 0.55034082
L = 13.04965918
L = 13.05
U = xbar + E
U = 13.6 + 0.55034082
U = 14.15034082
U = 14.15
Depending how you round the final answer, the 98% confidence interval for the population mean is approximately (13.05, 14.15)
This is in the format (L, U)
Another way to express the confidence interval would be to say 13.05 < mu < 14.15
This is of the format L < mu < U which I think is more descriptive since we're involving the parameter being estimated (mu).
|
|
|
