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 95% confidence interval is set up for the population standard deviation for the population of 1700 registered students
To determine the upper limit of the 95% confidence interval for the population standard deviation, we can use the Chi-Squared distribution.
Given:
Sample size (n) = 29
Sample mean (x̄) = 72.8%
Sample variance (s^2) = 275.56%
First, we need to calculate the chi-squared statistic.
The chi-squared statistic is given by:
χ² = (n-1)s^2 / σ²
Where σ is the population standard deviation.
Since we want an upper limit, we will use the upper tail of the chi-squared distribution.
For a 95% confidence interval, the critical chi-squared value for an upper tail probability of 0.05 is χ² = χ²_(0.05, n-1).
Now, we need to find χ²_(0.05, n-1). We can use a chi-squared table or chi-squared calculator to find this value.
Assuming n-1 = 28 (29-1), look up the chi-squared value for an upper tail probability of 0.05 and 28 degrees of freedom. The calculated value is χ²_(0.05, 28) = 43.773.
Plug in the values into the chi-squared statistic equation:
43.773 = (29-1)s^2 / σ²
Now, we need to rearrange the equation to solve for the upper limit of the population standard deviation (σ).
σ² = (29-1)s^2 / 43.773
Take the square root of both sides to get the upper limit of σ:
σ ≤ √((29-1)s^2 / 43.773)
Now, substitute the given sample variance:
σ ≤ √((29-1) * 275.56 / 43.773)
σ ≤ √(28 * 275.56 / 43.773)
Finally, calculate the upper limit of σ:
σ ≤ √(17367.68 / 43.773)
σ ≤ √397.106
σ ≤ 19.927
Therefore, the upper limit of the 95% confidence interval for the population standard deviation is approximately 19.927.
To determine the upper limit for a 95% confidence interval for the population standard deviation, we can use the chi-square distribution.
Given the sample size (n = 29), the sample variance (s^2 = 275.56%), and the sample mean, we can calculate the chi-square statistic.
The chi-square statistic is given by the formula:
χ^2 = (n - 1) * s^2 / σ^2
where n is the sample size, s^2 is the sample variance, and σ^2 is the population variance.
In this case, we want to find the upper limit of the 95% confidence interval, which corresponds to a significance level of α = 0.05. We need to find the chi-square value from the chi-square distribution table with a degrees of freedom (df) equal to n - 1.
The critical chi-square value for a 95% confidence interval and df = n - 1 can be found by subtracting α from 1 and looking up the value in the chi-square distribution table. The degrees of freedom for this scenario would be 28.
Let's calculate the upper limit for the 95% confidence interval:
1. Look up the critical chi-square value for a significance level of α = 0.05 and df = 28. In the chi-square distribution table, the value is approximately 45.0067.
2. Set up the equation:
χ^2 = (n - 1) * s^2 / σ^2
Substituting the values:
45.0067 = (29 - 1) * 275.56% / σ^2
3. Rearrange the equation to solve for σ:
σ^2 = (29 - 1) * 275.56% / 45.0067
4. Calculate the standard deviation σ:
σ = √[(29 - 1) * 275.56% / 45.0067]
σ ≈ √[28 * 2.7556 / 45.0067]
σ ≈ √(156.19728 / 45.0067)
σ ≈ √3.4717
σ ≈ 1.863
So, the upper limit for the 95% confidence interval for the population standard deviation is approximately 1.863.
To determine the upper limit of the 95% confidence interval for the population standard deviation, we need to use the chi-square distribution and the given sample statistics.
The chi-square distribution is commonly used for constructing confidence intervals for the population standard deviation when the underlying distribution is assumed to be approximately normal.
First, we need to calculate the degrees of freedom (df) for the chi-square distribution. In this case, since we have a sample size of 29, the degrees of freedom is given by (n - 1), where n is the sample size. Therefore, df = 28.
The critical values for the chi-square distribution depend on the confidence level and the degrees of freedom. For a 95% confidence level and df = 28, the critical value is approximately 45.722 (obtained from statistical tables or software).
Next, we need to calculate the chi-square value based on the given variance. The chi-square value is calculated as (n - 1) times the sample variance divided by the population variance. In this case, the sample variance is 275.56% and the population variance is unknown. Therefore, we use this sample variance as an estimate of the population variance. Substituting the values, we get:
chi-square = (28 * 275.56) / population variance
Next, we equate the chi-square value to the critical value and solve for the population variance:
(28 * 275.56) / population variance = 45.722
Simplifying the equation:
population variance = (28 * 275.56) / 45.722
Finally, we take the square root of the population variance to get the population standard deviation:
population standard deviation = √(population variance)
This will give us the upper limit of the 95% confidence interval for the population standard deviation for the population of 1700 registered students.