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
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.
To determine the lower limit of a 99% confidence interval for the population standard deviation, we can use the chi-square distribution. The formula for the confidence interval is:
Lower limit = [(n - 1) * s^2] / chi^2(alpha/2, n - 1)
Where:
- n = number of observations (in this case, 29)
- s^2 = sample variance (275.56%)
- chi^2(alpha/2, n - 1) = chi-square value corresponding to the desired significance level (in this case, alpha/2 = 0.01/2 = 0.005, and n - 1 = 28)
Using a chi-square table or calculator, we can find the chi-square value for alpha/2 = 0.005 and n - 1 = 28 to be approximately 49.888.
Substituting the values into the formula:
Lower limit = [(29 - 1) * 275.56%] / 49.888
Lower limit = (28 * 275.56%) / 49.888
Lower limit = (7716.08%) / 49.888
Lower limit ≈ 154,38
Therefore, the lower limit of the 99% confidence interval for the population standard deviation is approximately 154.38.
To determine the lower limit for a 99% confidence interval for the population standard deviation, we need to use the chi-square distribution.
The formula for the confidence interval is:
[ ((n - 1) * s^2) / χ^2_upper , ((n - 1) * s^2) / χ^2_lower ]
Where:
- n = sample size (29 in this case)
- s^2 = variance (275.56% in this case)
- χ^2_upper = chi-square value for the upper limit (we need to find this)
- χ^2_lower = chi-square value for the lower limit (this is the value we want to find)
Since we want a 99% confidence interval, we need to find the chi-square values that correspond to the upper and lower tails of 0.005 (0.01 / 2) each.
Using a chi-square table (or a calculator), we find that the chi-square value for the upper tail probability of 0.005 with 28 degrees of freedom (n-1) is approximately 48.216.
To find the lower limit, we need to find the chi-square value for the lower tail probability of 0.005 with 28 degrees of freedom. Since the chi-square distribution is symmetric, this is the same as the upper tail probability of 0.995.
Using a chi-square table (or a calculator), we find that the chi-square value for the lower tail probability of 0.995 with 28 degrees of freedom is approximately 12.401.
Now we can substitute these values into the formula:
[ ((29 - 1) * 275.56%) / 48.216 , ((29 - 1) * 275.56%) / 12.401 ]
Simplifying the expression:
[ 28 * 275.56% / 48.216 , 28 * 275.56% / 12.401 ]
Converting percentages to decimals:
[ 28 * 2.7556 / 48.216 , 28 * 2.7556 / 12.401 ]
Calculating the values:
[ 1.597 , 3.556 ]
Therefore, the lower limit for the 99% confidence interval for the population standard deviation is approximately 1.597.