Find the 90% confidence interval for the variance and the standard deviation of the ages of seniors at Oak Park College if a sample of 24 students has a standard deviation of 2,3 years. Assume the variable is normally distribution
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± 1.645 SEm
Standard Error of the mean = SEm = SD/√(n-1)
That is for a distribution of means.
For a distribution of scores
± 1.645 SD
Given that the distance from the earth to the sun is 1.5 times 10 to the power m and the
To find the 90% confidence interval for the variance and the standard deviation, we can use the Chi-Square distribution.
Step 1: Calculate the Chi-Square critical values.
Since we have a 90% confidence level, we need to find the critical values from the Chi-Square distribution with degrees of freedom (df) equal to the sample size minus 1.
Degrees of Freedom (df) = n - 1 = 24 - 1 = 23
Using a Chi-Square table or a statistical software, find the critical values Chi-Square (χ²) corresponding to a 90% confidence level and 23 degrees of freedom. Let's call these values χ²₁ and χ²₂.
Step 2: Calculate the confidence interval for variance.
The confidence interval for the variance can be calculated using the formula:
[ (n - 1) * s² / χ²₂ , (n - 1) * s² / χ²₁ ]
where n is the sample size and s² is the sample variance.
In this case, n = 24 and s² = (standard deviation)^2 = (2.3)^2 = 5.29.
Plug in the values to get the confidence interval for the variance.
[ (24 - 1) * 5.29 / χ²₂ , (24 - 1) * 5.29 / χ²₁ ]
Step 3: Calculate the confidence interval for standard deviation.
The confidence interval for the standard deviation can be calculated by taking the square root of the confidence interval for the variance:
[ √( (n - 1) * s² / χ²₂ ) , √( (n - 1) * s² / χ²₁ ) ]
Plug in the values from Step 2 to get the confidence interval for the standard deviation.
[ √( (24 - 1) * 5.29 / χ²₂ ) , √( (24 - 1) * 5.29 / χ²₁ ) ]
Note: The critical values χ²₁ and χ²₂ depend on the desired confidence level and degrees of freedom. Make sure to use the correct values for your specific case.