SOLUTION: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A rando

Question 1201399: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A random sample of 15 AAA batteries produced by this manufacturer lasted a mean of 9.5 hours with a standard deviation of 1.7 hours. Find a 95% confidence interval for the population standard deviation of the lifetimes of AAA batteries produced by the manufacturer. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
Found 2 solutions by Theo, math_tutor2020:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sample mean is 9.5 hours.
sample standard devition is 1.7 hours.
sample size is 15.
t-score is indicated because the standard deviation is taken from the sample rather than the population.
since you are looking at the mean of a sample of several elements, you would use the standard error rather than the standard deviation.
standard error = standard deviation / sqrt(sample size) = 1.7 / sqrt(15) = .4389381126.
t-score formula is t = (x - m) / s
t is the t-score
x is the upper or lower limit of the 95% considence interval.
m is mean of the sample.
s is the standard error.
the 95% confidence interval t-tscore with 14 degrees of freedom is plus or minus 2.144786681.
for the upper limit of the confidence interval, the t-score formula becomes:
2.144786681 = (x - 9.5) / .4389381126
solve for x to get:
x = 2.144786681 * .4389381126 + 9.5 = 10.44142862.
for the lower limit of the confidence interval, the t-score formula becomes:
-2.144786681 = (x - 9.5) / .4389381126
solve for x to get:
x = -2.144786681 * .4389381126 + 9.5 = 8.558571382.
your 95% confidence interval is 8.558571382 to 10.44142862.
round to two decimal places to get:
your 95% confidence interval is 8.56 to 10.44.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

sigma = population standard deviation

Goal: estimate sigma

This estimate is written as a confidence interval in the format
L < sigma < U
where,
L = lower boundary
U = upper boundary

s = sample standard deviation, which helps estimate sigma
s = 1.7 is given in the instructions

n = sample size = 15
df = degrees of freedom
df = n-1
df = 15-1
df = 14

Here is an article to check out
https://faculty.elgin.edu/dkernler/statistics/ch09/9-3.html

Check out example 1 to see how to determine the critical values based on the degrees of freedom (df) and on the confidence level.
Example 1 uses df = 12, but the confidence level used is 95%. Meaning that you'll look at those same columns to get the XL and XR values (except just look at the row df = 14 instead)

If you were to look at the df = 14 row, and those columns mentioned, then you should find
XL = 5.629 = left critical chi-square value
XR = 26.119 = right critical chi-square value

This would indicate
P(XL < X2 < XR) = 0.95
P(5.629 < X2 < 26.119) = 0.95
The area under the chi-square curve, between 5.629 and 26.119, is roughly 0.95
About 95% of the area under the curve is between 5.629 and 26.119

Side note: A chi-square calculator can compute the left and right critical values needed.

Now we can compute the lower boundary of the confidence interval for sigma.

Now compute the upper boundary.

It could be a bit confusing that XR goes for the left or lower boundary, while XL goes for the right or upper boundary.

The confidence interval format
L < sigma < U
is then updated to
1.245 < sigma < 2.681
which is approximate.

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