Answer :
The 95% confidence interval for the population variance is [0.02, 0.23].
Calculating the confidence interval for the population variance?
From the question, we have the following parameters that can be used in our computation:
The table of values
Calculate the sample mean and sample variance as follows
Sample mean = (1.74 + 1.84 + 1.53 + 1.62 + 1.78 + 1.95 + 1.37 + 1.55 + 1.42 + 2.04) / 10 = 1.68
Sample variance = (0.003136 + 0.024336 + 0.023716 + 0.004096 + 0.009216 + 0.070756 + 0.098596 + 0.017956 + 0.069696 + 0.126736) / 10 = 0.044824
Calculate the degrees of freedom.
Degrees of freedom = n - 1
df = 10 - 1 = 9
For a 95% confidence interval, the chi-squared values are 1.7345 and 16.9190.
So, we have
Limit = (degrees of freedom * sample variance) / chi-squared value
This gives
Lower limit = (9 * 0.044824) / 16.9190 = 0.02
Upper limit = (9 * 0.044824) / 1.7345 = 0.23
Hence, the 95% confidence interval for the population variance is [0.02, 0.23].
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Question
The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown to the right. Assume the sample is taken from a normally distributed population.
What is the confidence interval for the population variance? (Round to two decimal places as needed)
X (X-Xbar)^2
1.74 0.003136
1.84 0.024336
1.53 0.023716
1.62 0.004096
1.78 0.009216
1.95 0.070756
1.37 0.098596
1.55 0.017956
1.42 0.069696
2.04 0.126736
Total 16.84 0.44824