Answer:
Lower endpoint: 0.31 , Upper endpoint: 0.79
Explanation:
In the problem statement we are told c=98% (or 0.98) and n=54, so we know α=1–0.98=0.02 and df=54–1=53.
Note, 53 degrees of freedom is not in the chi-square distribution table, so we can either approximate χ20.01 and χ20.99 using the closest given value or we can use statistical software to find more exact values.
Using the chi-square distribution table, approximating the degrees of freedom as 50, we find that the critical values are 76.154 and 29.707. These are the critical values we will use to construct the confidence interval, but alternatively, we could have used technology to find the critical values χ20.01=79.843 and χ20.99=32.018.
A frozen food company uses a machine that packages okra in six-ounce portions. A sample of 54 packages of okra has a variance of 0.44. Construct the 98% confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two decimal places.
Finally, we substitute these values into the formula for the confidence interval for population variance and simplify:
(n−1)s2χ2α/2<σ2<(n−1)s2χ2(1−α/2)
(54−1)0.4476.154<σ2<(54−1)0.4429.707
0.31<σ2<0.79
Lower endpoint: 0.31, Upper endpoint: 0.79