SOLUTION: The following data are a random sample of turtles and their weights (ounces).
{18,16,20,12,11,10,16,17}
Assume the underlying population distribution is approximately normal and
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Question 1198875: The following data are a random sample of turtles and their weights (ounces).
{18,16,20,12,11,10,16,17}
Assume the underlying population distribution is approximately normal and has no outliers. Find the following, rounding your intermediate steps and answers to 3 decimal places.
a. Find the sample standard deviation.
s=
b. Find the 95% confidence interval for the population standard deviation.
(,)
c. Find the 99% confidence interval for the population standard deviation.
(,)
d. As expected, the answer for part c. is ____ than the answer for part b.
e. The lower bound is ____ the upper bound is from s because of the shape of the χ2 distribution.
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**a. Sample Standard Deviation**
1. **Calculate the sample mean (x̄):**
x̄ = (18 + 16 + 20 + 12 + 11 + 10 + 16 + 17) / 8 = 15
2. **Calculate the squared differences from the mean:**
(18-15)² = 9
(16-15)² = 1
(20-15)² = 25
(12-15)² = 9
(11-15)² = 16
(10-15)² = 25
(16-15)² = 1
(17-15)² = 4
3. **Sum the squared differences:**
9 + 1 + 25 + 9 + 16 + 25 + 1 + 4 = 89
4. **Calculate the sample variance (s²):**
s² = Σ(x - x̄)² / (n - 1) = 89 / (8 - 1) = 12.714
5. **Calculate the sample standard deviation (s):**
s = √s² = √12.714 = 3.566
**Therefore, the sample standard deviation (s) is 3.566**
**b. 95% Confidence Interval for the Population Standard Deviation**
* **Find the chi-square values:**
* Degrees of freedom (df) = n - 1 = 8 - 1 = 7
* Using a chi-square table or calculator:
* χ²_lower (for 0.025 area in the right tail) = 2.167
* χ²_upper (for 0.025 area in the left tail) = 18.475
* **Calculate the confidence interval:**
* Lower bound: √[(n - 1) * s² / χ²_upper] = √[(7 * 12.714) / 18.475] = 2.172
* Upper bound: √[(n - 1) * s² / χ²_lower] = √[(7 * 12.714) / 2.167] = 6.390
* **95% Confidence Interval: (2.172, 6.390)**
**c. 99% Confidence Interval for the Population Standard Deviation**
* **Find the chi-square values:**
* Degrees of freedom (df) = 7
* Using a chi-square table or calculator:
* χ²_lower (for 0.005 area in the right tail) = 0.989
* χ²_upper (for 0.005 area in the left tail) = 20.278
* **Calculate the confidence interval:**
* Lower bound: √[(7 * 12.714) / 20.278] = 1.974
* Upper bound: √[(7 * 12.714) / 0.989] = 9.446
* **99% Confidence Interval: (1.974, 9.446)**
**d. As expected, the answer for part c. is wider** than the answer for part b.
**e. The lower bound is lower** and **the upper bound is higher** from s because of the shape of the chi-square distribution.
* The chi-square distribution is skewed to the right.
* For a higher confidence level (like 99%), we need a wider interval to capture more of the possible values of the population standard deviation. This results in a lower lower bound and a higher upper bound compared to the 95% confidence interval.
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