Asked by Sydni Redding on Jun 14, 2024
Verified
A random sample of 31 sales charge showed a sample standard deviation of $50.A 90% confidence interval estimate of the population standard deviation is
A) 1715.101 to 4055.589.
B) 1596.458 to 4466.679.
C) 39.956 to 66.833.
D) 41.393 to 63.684.
Standard Deviation
Indicates how dispersed the values in a data set are, relative to the mean of the set.
- Calculate and interpret confidence intervals for population variance and standard deviation.
Verified Answer
JD
Jennifer DeleonJun 17, 2024
Final Answer :
D
Explanation :
We use the chi-square distribution with (n-1) degrees of freedom to construct a confidence interval for population standard deviation as follows:
((n-1)*s^2)/X^2 ≤ σ^2 ≤ ((n-1)*s^2)/X^2
where X^2 is the chi-square value for the given confidence level and (n-1) degrees of freedom.
For this problem, n=31 and the sample standard deviation s=$50. Using a chi-square distribution calculator with 30 degrees of freedom and a 90% confidence level, we get X^2=42.557. Substituting the values, we get the confidence interval as:
(30*50^2)/63.69 ≤ σ^2 ≤ (30*50^2)/26.51
or 41.393 ≤ σ ≤ 63.684 (rounded to three decimal places). Therefore, the answer is D.
((n-1)*s^2)/X^2 ≤ σ^2 ≤ ((n-1)*s^2)/X^2
where X^2 is the chi-square value for the given confidence level and (n-1) degrees of freedom.
For this problem, n=31 and the sample standard deviation s=$50. Using a chi-square distribution calculator with 30 degrees of freedom and a 90% confidence level, we get X^2=42.557. Substituting the values, we get the confidence interval as:
(30*50^2)/63.69 ≤ σ^2 ≤ (30*50^2)/26.51
or 41.393 ≤ σ ≤ 63.684 (rounded to three decimal places). Therefore, the answer is D.
Learning Objectives
- Calculate and interpret confidence intervals for population variance and standard deviation.