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.
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
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