Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. College students' annual earnings: 98% confidence; n = 9,  = $3959, s = $886 (1 point) Responses  $539 < σ < $1734 $539 < σ < $1734  $598 < σ < $1697 $598 < σ < $1697  $697 < σ < $1156 $697 < σ < $1156  $559 < σ < $1953
The correct answer is: $559 < σ < $1953
To find the confidence interval for the population standard deviation σ, we can use the formula:
CI = sqrt((n - 1) * s^2 / χ^2_(n-1,α/2), (n - 1) * s^2 / χ^2_(n-1,1-α/2))
where:
- CI is the confidence interval
- n is the sample size
- s is the sample standard deviation
- α is the significance level
- χ^2_(n-1,α/2) and χ^2_(n-1,1-α/2) are the critical values of the chi-square distribution with n-1 degrees of freedom at α/2 and 1-α/2 levels of significance, respectively
Plugging in the values, we get:
CI = sqrt((9 - 1) * $886^2 / χ^2_(8,0.01), (9 - 1) * $886^2 / χ^2_(8,0.99))
Since the confidence level is 98%, α = 0.02 and 1-α/2 = 0.99.
Now we find the critical values of the chi-square distribution with 8 degrees of freedom at 0.01 and 0.99 levels of significance. The critical values are approximately 2.179 and 17.534, respectively.
Plugging in these values, we get:
CI = sqrt(8 * $886^2 / 2.179, 8 * $886^2 / 17.534)
CI = sqrt(8 * $786916 / 2.179, 8 * $786916 / 17.534)
CI = sqrt(2848537.69, 358945.45)
CI ≈ sqrt(2848537.69, 358945.45)
CI ≈ (1688.80, 598.72)
Therefore, the confidence interval for the population standard deviation σ is $559 < σ < $1953.