Question 1039782: A food snack manufacturer samples 9 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.4 and the sample standard deviation is 0.15, find the 95% confidence interval of the true mean.
A.
(10.30, 10.50)
B.
(10.05, 10.75)
C.
(10.35, 10.45)
D.
(10.28, 10.52)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Use a table like this one to find the t critical value is t = 2.306
How to find this value? Look at the row that has df = 8 and look at the column that corresponds to 95% confidence. The value 2.306 is found at this row and column intersection
Note: degrees of freedom = df = n-1 = 9-1 = 8
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We have the following information
xbar = 10.4 is the sample mean
s = 0.15 is the standard deviation
n = 9 is the sample size
t = 2.306 is the t critical value (found above)
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Let's compute the lower and upper boundaries
L = lower boundary
U = upper boundary
L = xbar - t*s/sqrt(n)
L = 10.4 - 2.306*0.15/sqrt(9)
L = 10.2847
L = 10.28
U = xbar + t*s/sqrt(n)
U = 10.4 + 2.306*0.15/sqrt(9)
U = 10.5153
U = 10.52
The 95% confidence interval, for the population mean mu, is therefore (L,U) = (10.28, 10.52) which is choice D
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