To calculate the upper limit of a 99% confidence interval for the population variance, we will use the chi-square distribution.
The chi-square distribution has degrees of freedom equal to the sample size minus 1, which in this case is 29 - 1 = 28.
We need to find the value of chi-square that corresponds to a cumulative probability of 0.01 (1% significance level) with 28 degrees of freedom.
Using a chi-square table or calculator, we find the value of chi-square to be approximately 52.29.
The upper limit of the 99% confidence interval for the population variance can be calculated as:
upper limit = (n - 1) * sample variance / chi-square
Where n = total sample size and sample variance = variance of the sample.
In this case, n = 1700 (population size) and sample variance = 275.56%.
upper limit = (1700 - 1) * 275.56% / 52.29
= 1699 * 2.7556 / 52.29
= 89,823.44 / 52.29
≈ 1,718.25
Therefore, the upper limit for the 99% confidence interval for the population variance is approximately 1,718.25.
None of the provided options match this value.
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
619.24%
147.41%
173.54%
151.32%
503.96%
1 answer
1 answer