Question 1058636: The following data were randomly drawn from an approximately normal population.
2, 6, 11, 12, 21, 23
use these data, find a 95% confidence interval for the population standard deviation. Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
what is the lower limit of the 95% confident interval and what is the upper limit of the 95% confident interval
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The following data were randomly drawn from an approximately normal population.
2, 6, 11, 12, 21, 23
Sample Variance = 8.22^2 = 67.568
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use these data, find a 95% confidence interval for the population standard deviation.
(n-1)s^2/Xr < population variance < (n-1)s^2/Xl
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5*67.568/11.143 < p-variance < 5*67.568/0.484
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Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
Lower limit of the 95% CI:: 30.319
Upper limit of the 95% CI:: 698.017
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Cheers,
Stan H.
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Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! Since we are trying to estimate the standard deviation in the population, we choose the standard deviation as the sample statistic.
:
Sample mean is (2 + 6 + 11 + 12 + 21 + 23) / 6 = 12.5
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Sample standard deviation is square root( (1/6) * summation from 1 to 6 (x(i) - sample mean)^2 ) = 8.2158
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degrees of freedom(df) = 6 - 1 = 5
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alpha(a) = 1 - (95/100) = 0.005
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critical probability(p*) = 1 - (a/2) = 0.975
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we use the chi-square table to find two values
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x^2 (1 - (a/2) ) and 5 degrees of freedom = 0.831
x^2(a/2) and five degrees of freedom = 12.833
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the 95% confidence interval for the variance is
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5*(8.2158)^2/12.833 < variance < 5*(8.2158)^2/0.831
:
26.2991 < variance < 406.1334
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standard deviation = square root of the variance
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95% confidence interval for the population standard deviation
:
(5.1283, 20.1528)
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