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Midterm Exam 2
Question 4 of 8 (1 point)
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Question 4
The scores on an examination in biology are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination.
435, 453, 473, 537, 559, 603

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Find a 90% confidence interval for the population standard deviation. Then give its lower limit and upper limit.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

Lower limit:
Upper limit:

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To find a 90% confidence interval for the population standard deviation based on the sample provided, we can use the Chi-square distribution. Here's a step-by-step guide on how to calculate the confidence interval:

### Step 1: List the given data
- Sample dataTo find a 90% confidence interval for the population standard deviation based on the sample provided, we can use the Chi-square distribution. Here's a step-by-step guide on how to calculate the confidence interval:

### Step 1: List the given data
- Sample  435, 453, 473, 537, 559, 603
- Mean (\(\mu\)) = 500
- Confidence level = 90%
- Sample size (\(n\)) = 6

### Step 2: Calculate the sample variance
The sample variance (\(s^2\)) is given by:

\[
s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2
\]

where \( \bar{x} \) is the sample mean, and \( x_i \) are the individual data points.

1. Compute the sample mean (\(\bar{x}\)):

\[
\bar{x} = \frac{435 + 453 + 473 + 537 + 559 + 603}{6} = 510
\]

2. Compute the squared deviations from the mean and sum them up:

\[
(435 - 510)^2 = 5625
\]
\[
(453 - 510)^2 = 3249
\]
\[
(473 - 510)^2 = 1369
\]
\[
(537 - 510)^2 = 729
\]
\[
(559 - 510)^2 = 2401
\]
\[
(603 - 510)^2 = 8649
\]

\[
\text{Sum of squared deviations} = 5625 + 3249 + 1369 + 729 + 2401 + 8649 = 22022
\]

3. Calculate the sample variance:

\[
s^2 = \frac{22022}{6-1} = \frac{22022}{5} = 4404.4
\]

4. The sample standard deviation (\(s\)) is:

\[
s = \sqrt{4404.4} = 66.37
\]

### Step 3: Use the Chi-square distribution
The confidence interval for the population variance is calculated using the formula:

\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right)
\]

For a 90% confidence interval, \(\alpha = 0.10\), so:

\[
\alpha/2 = 0.05
\]

Using a Chi-square table with \( n-1 = 5 \) degrees of freedom:

- \(\chi^2_{0.05}\) = 11.070
- \(\chi^2_{0.95}\) = 0.831

### Step 4: Calculate the confidence interval for variance
Now, substitute into the formula for variance:

Lower limit for variance:

\[
\frac{(6-1) \times 4404.4}{11.070} = \frac{5 \times 4404.4}{11.070} = \frac{22022}{11.070} = 1989.27
\]

Upper limit for variance:

\[
\frac{(6-1) \times 4404.4}{0.831} = \frac{22022}{0.831} = 26500.60
\]

### Step 5: Confidence interval for the standard deviation
To find the confidence interval for the standard deviation, take the square root of both limits:

- Lower limit: \(\sqrt{1989.27} = 44.60\)
- Upper limit: \(\sqrt{26500.60} = 162.77\)

### Final Answer:
- Lower limit: **44.60**
- Upper limit: **162.77**

Would you like more details or have any questions? Here are some related questions to consider:

1. What is the role of the Chi-square distribution in calculating confidence intervals?
2. How does the sample size affect the confidence interval for the standard deviation?
3. What happens to the confidence interval if the confidence level is increased?
4. Can the confidence interval for the population mean also be calculated from the same data?
5. How do you interpret a 90% confidence interval in practice?

**Tip:** When working with confidence intervals, always ensure that your degrees of freedom and significance level match the appropriate distribution table.

The scores on an examination in biology are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination: 435, 453, 473, 537, 559, 603. Find a 90% confidence interval for the population standard deviation. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

Calculate the 90% confidence interval for the population standard deviation based on a random sample of biology exam scores: 435, 453, 473, 537, 559, 603. Learn how to use the Chi-square distribution, with intermediate steps shown and results rounded to two decimal places.

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