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
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.
To determine the upper limit for the population variance with a 99% confidence interval, we will use the chi-square distribution.
The chi-square statistic is calculated as follows:
χ^2 = (n-1) * s^2 / σ^2
where:
χ^2 = chi-square statistic
n = sample size
s^2 = sample variance
σ^2 = population variance
Given:
Sample size (n) = 29
Sample variance (s^2) = 275.56%
Mean of the sample = 72.8%
Population size (N) = 1700
First, let's find the chi-square critical value for a 99% confidence interval with degrees of freedom (df) = n - 1.
df = 29 - 1 = 28
Using a chi-square table or a statistical software, the critical value for a 99% confidence interval and 28 degrees of freedom is approximately 44.314.
Next, we need to solve the chi-square statistic formula for the population variance (σ^2):
χ^2 = (n-1) * s^2 / σ^2
Rearranging the formula:
σ^2 = (n-1) * s^2 / χ^2
Substituting the given values into the formula:
σ^2 = (29-1) * 275.56% / 44.314
Calculating:
σ^2 ≈ 1689.55
Therefore, the upper limit for the population variance, with a 99% confidence interval, is approximately 1689.55.
To determine the upper limit for a 99% confidence interval for the population variance, we can use the chi-square distribution and the given sample statistics.
The chi-square distribution is commonly used for hypothesis testing and confidence interval estimation for population variances. The upper limit for a confidence interval can be found using the chi-square distribution's upper critical value.
First, let's calculate the degrees of freedom (df). In this case, the degrees of freedom is equal to the sample size minus 1:
df = n - 1 = 29 - 1 = 28
The upper critical value for the chi-square distribution can be found using the chi-square table or statistical software. For a 99% confidence level, the upper critical value is the value that leaves a 1% area in the upper tail of the chi-square distribution with df degrees of freedom.
In this case, the upper critical value for df = 28 and a 1% area in the upper tail is approximately 48.9955.
Next, we can calculate the upper limit of the confidence interval for the population variance using the formula:
Upper Limit = ((n - 1) * sample_variance) / critical_value
Using the given sample variance of 275.56%, we can substitute the values into the formula:
Upper Limit = ((29 - 1) * 275.56) / 48.9955
Upper Limit = 28 * 275.56 / 48.9955
Upper Limit = 1562.08
Therefore, the upper limit for the 99% confidence interval for the population variance of the 1700 registered students is 1562.08.