Question 1171999: A sociologist develops a test to measure attitudes about public transportation, and 27
randomly selected subjects are given the test. Their mean score is 76.2 and their standard
deviation is 21.4. Construct and interpret the 95% confidence interval for the standard
deviation, o, of the scores of all subjects.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down how to construct and interpret the 95% confidence interval for the population standard deviation (σ) of the test scores.
**1. Identify the Given Information**
* Sample size (n) = 27
* Sample mean (x̄) = 76.2 (This is not needed for the standard deviation CI)
* Sample standard deviation (s) = 21.4
* Confidence level = 95%
**2. Determine the Degrees of Freedom**
* Degrees of freedom (df) = n - 1 = 27 - 1 = 26
**3. Find the Chi-Square Critical Values**
* We need to find the chi-square (χ²) values that correspond to the lower and upper tails of the distribution for a 95% confidence interval.
* Since it's a 95% confidence interval, the area in each tail is (1 - 0.95) / 2 = 0.025.
* We will use a chi-square distribution table or a calculator/software to find the critical values.
* χ²_lower (df=26, area=0.975) ≈ 13.844
* χ²_upper (df=26, area=0.025) ≈ 41.923
**4. Calculate the Confidence Interval**
* The formula for the confidence interval for the population standard deviation (σ) is:
```
sqrt[ ( (n - 1) * s² ) / χ²_upper ] < σ < sqrt[ ( (n - 1) * s² ) / χ²_lower ]
```
* Plug in the values:
```
sqrt[ ( 26 * (21.4)² ) / 41.923 ] < σ < sqrt[ ( 26 * (21.4)² ) / 13.844 ]
```
* Calculate:
* (n-1)*s^2 = 26 * (21.4)^2 = 11883.44
* 11883.44 / 41.923 = 283.456. sqrt(283.456) = 16.836
* 11883.44 / 13.844 = 858.382. sqrt(858.382) = 29.298
* Therefore:
```
16.836 < σ < 29.298
```
**5. Interpret the Confidence Interval**
* We are 95% confident that the true population standard deviation of attitudes about public transportation scores lies between 16.836 and 29.298.
* In practical terms, this means that while the sample standard deviation was 21.4, the variability in the entire population's scores could reasonably be as low as about 16.8 or as high as about 29.3.
**Summary**
The 95% confidence interval for the population standard deviation of attitudes about public transportation is (16.836, 29.298). This indicates that the spread of attitude scores in the entire population is likely within this range.
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