Question 1161897: The lengths (in minutes) of a random selection of popular children’s animated films are listed below. Estimate the true mean length of all children’s animated films with 95% confidence. Assume the variable is normally distributed.
93 83 76 92 77 81 78 100 78 76 75
Sample = 11
Mean = 82.64
Confid =95% = 1.96
SD = 8.49
76.9 to 88.3 is the correct answer
77.6 to 87.6 is the answer that I am getting
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
At 95% confidence, the z value z = 1.96 (approx) is only applicable if we know the population standard deviation (sigma), or if n > 30. The value of sigma is not given and n = 11 is not greater than 30. So we cannot use the Z distribution.
Instead, we use a T distribution, with
df = degrees of freedom
df = n-1
df = 11-1
df = 10
We need to find the value of k such that P(-k < T < k) = 0.95 under a T distribution with df = 10. Using a T table, such as the one found here
http://www.ttable.org/
we find the value of k is roughly k = 2.228
Note: locate the row that has df = 10 at the beginning, then locate the column that has "two tails = 0.05". The two tails combine in area of 0.05 with the body being 0.95 to represent a 95% confidence level. At this row and column of the table is 2.228
So because P(-2.228 < T < 2.228) = 0.95 approximately, when df = 10, we can say the critical t value is t = 2.228 approximately.
Therefore, the lower (L) and upper (U) bounds of the confidence interval are
L = xbar - t*s/sqrt(n)
L = 82.64 - 2.228*8.49/sqrt(11)
L = 76.93 approximately
L = 76.9
and
U = xbar + t*s/sqrt(n)
U = 82.64 + 2.228*8.49/sqrt(11)
U = 88.34 approximately
U = 88.3
The 95% confidence interval is roughly (L,U) = (76.9, 88.3)
we can write this in the form L < mu < U to get 76.9 < mu < 88.3
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