A random sample of 49 observations is used to estimate the population variance.

Question

A random sample of 49 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 54.5 and 3.3, respectively. Assume that the population is normally distributed. Use Table 3.

Construct a 90% interval estimate of the population variance. (Do not round intermediate calculations. Round your answers to 2 decimal places.)

Construct a 99% interval estimate of the population variance. (Do not round intermediate calculations. Round your answers to 2 decimal places.)

A random sample of 49 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 54.5 and 3.3, respectively. Assume that the population is normally distributed. Use Table 3.

Explanation / Answer

a.
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S^2 = Variance
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.1
^2 right = (1 - Confidence Level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 48 df are 65.1708 , 33.098
Variacne( S^2 )=10.89
Sample Size(n)=49
Confidence Interval = [ 48 * 10.89/65.1708 < ^2 < 48 * 10.89/33.098 ]
= [ 522.72/65.1708 < ^2 < 522.72/33.0981 ]
[ 8.0208 , 15.7931 ]

b.
AT 99%
Confidence Interval = [ 48 * 10.89/76.9688 < ^2 < 48 * 10.89/26.511 ]
= [ 522.72/76.9688 < ^2 < 522.72/26.5106 ]
[ 6.7913 , 19.7174 ]

c.
The width of the interval increases with the confidence level.

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