Question 1202095: If n=17,¯x(x-bar)=34, and s=4, construct a confidence interval at a 99% confidence level.
Assume the data came from a normally distributed population.
Give your answers to one decimal place.
_____<μ<_____
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 31.2 < mu < 36.8
Work Shown:
mu = μ = Greek letter representing population mean
n = 17 = sample size
xbar = 34 = sample mean
s = 4 = sample standard deviation
df = degrees of freedom = n-1 = 17-1 = 16
Because the population standard deviation (sigma) is not known, and because n > 30 isn't the case, we must use the T distribution.
Refer to this T table
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Locate the row labeled df = 16
Locate the columnn labeled "99% confidence". The confidence labels are at the bottom.
The approximate value t = 2.921 is at this row and column intersection. It is the t critical value.
Specialized stats calculators such as this one
https://www.omnicalculator.com/statistics/critical-value
can find the t critical value.
Make sure to do a two-tailed test.
Also set the significance level to 0.01 (recall that alpha = 1-C where C is the confidence level)
Compute the margin of error
E = t*s/sqrt(n)
E = 2.921*4/sqrt(17)
E = 2.833786
That result is approximate.
Then,
L = lower boundary
L = xbar - E
L = 34 - 2.833786
L = 31.166214
L = 31.2
and
U = upper boundary
U = xbar + E
U = 34 + 2.833786
U = 36.833786
U = 36.8
The 99% confidence interval in the format L < mu < U is roughly 31.2 < mu < 36.8
That can be condensed to the format (L, U) so we get (31.2, 36.8)
We are 99% confident that the population mean (mu) is somewhere between 31.2 and 36.8
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