To determine the lower limit of the 99% confidence interval for the population standard deviation, we can use the chi-square distribution.
Given:
Sample size (n) = 29
Sample mean (x̄) = 72.8%
Sample variance (s^2) = 275.56%
Degrees of freedom (df) for the chi-square distribution = n - 1 = 29 - 1 = 28
Significance level (α) = 1 - Confidence level = 1 - 0.99 = 0.01
Using the chi-square distribution table or a chi-square calculator, find the chi-square value (χ^2) corresponding to a significance level of 0.01 and 28 degrees of freedom. The chi-square value should be 50.998.
The lower limit (LL) of the 99% confidence interval for the population standard deviation (σ) is given by the formula:
LL = sqrt((n - 1) * s^2 / χ^2)
Substituting the values into the formula:
LL = sqrt((29 - 1) * 275.56% / 50.998)
LL = sqrt(28 * 2.7556 / 50.998)
LL = sqrt(1.530416 / 50.998)
LL = sqrt(0.03000525)
LL ≈ 0.1731
Therefore, the lower limit of the 99% confidence interval for the population standard deviation is approximately 0.1731.
the test marks for 29 students studying STSA 1624 produced a mean of 72,8% and a variance of 275,56%. If the distribution of the marks may be assumed to be approximately normally distributed, determine the lower limit, if a 99% confidence interval is set up for the population standard deviation for the population of 1700 registered students
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