To determine the upper limit of the 99% confidence interval for the population variance, we need to use the chi-square distribution.
First, we need to calculate the chi-square statistic using the given sample variance and sample size:
chi-square = ((n-1) * sample variance) / population variance
chi-square = ((29-1) * 275.56) / population variance
Next, we need to find the critical chi-square value from the chi-square distribution table for a 99% confidence level with (n-1) degrees of freedom. Since there are 29 students in the sample, the degrees of freedom will be 29-1 = 28.
Using the chi-square distribution table, the critical chi-square value for a 99% confidence level with 28 degrees of freedom is approximately 45.722.
Now, we can set up the inequality to find the upper limit of the confidence interval for the population variance:
((n-1) * sample variance) / population variance < critical chi-square value
Plug in the values:
((29-1) * 275.56) / population variance < 45.722
Simplify the equation:
(28 * 275.56) / 45.722 < population variance
Multiply both sides by the population variance:
(28 * 275.56) / 45.722 * population variance < population variance^2
Simplify further:
167.5 < population variance
Therefore, the upper limit of the 99% confidence interval for the population variance is 167.5.
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 upper limit, if a 99% confidence interval is set up for the population variance for the population of 1700 registered students
1 answer
1 answer