To construct a 95% confidence interval for the population standard deviation $\sigma$ at Bank A, we need to follow these steps:

1. Calculate the sample standard deviation: First, compute the sample standard deviation $s$ from the given data at Bank A.

2. Degrees of freedom: The degrees of freedom $df$ for this calculation is $n - 1$, where $n$ is the number of data points.

3. Use the Chi-square distribution: The 95% confidence interval for the population standard deviation $\sigma$ is based on the Chi-square distribution. The formula for the confidence interval is:

   $\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{upper}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{lower}}} \right)$ where:

   • $\chi^2_{upper}$ and $\chi^2_{lower}$ are the critical values of the Chi-square distribution for the chosen confidence level.
   • $s^2$ is the sample variance.

Let's go through these steps.

Step 1: Calculate the sample standard deviation for Bank A

Data for Bank A:
6.5, 6.6, 6.7, 6.8, 7.1, 7.3, 7.6, 7.8, 7.8, 7.8

The sample mean $\bar{x}$ is calculated as:

$\bar{x} = \frac{6.5 + 6.6 + 6.7 + 6.8 + 7.1 + 7.3 + 7.6 + 7.8 + 7.8 + 7.8}{10} = 7.2$

Now, the sample standard deviation $s$ is:

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

$s = \sqrt{\frac{(6.5 - 7.2)^2 + (6.6 - 7.2)^2 + \dots + (7.8 - 7.2)^2}{9}} = 0.53$

Step 2: Degrees of freedom

The degrees of freedom $df$ is $n - 1 = 10 - 1 = 9$.

Step 3: Use the Chi-square distribution

From the table of Chi-square critical values for a 95% confidence interval and $df = 9$:

• $\chi^2_{upper} = 16.919$
• $\chi^2_{lower} = 3.325$

Step 4: Calculate the confidence interval

Now we can calculate the confidence interval for the population standard deviation $\sigma$ as follows:

$\text{Lower bound} = \sqrt{\frac{(9)(0.53^2)}{16.919}} = \sqrt{\frac{2.53}{16.919}} = \sqrt{0.1496} = 0.39$

$\text{Upper bound} = \sqrt{\frac{(9)(0.53^2)}{3.325}} = \sqrt{\frac{2.53}{3.325}} = \sqrt{0.761} = 0.87$

Final Answer:

The 95% confidence interval for the population standard deviation $\sigma$ at Bank A is:

$0.39 \, \text{min} < \sigma_A < 0.87 \, \text{min}$

Would you like further details or explanations? Here are a few related questions to explore:

1. How do you interpret the confidence interval for standard deviation?
2. How would the confidence interval change if the sample size were larger?
3. Why do we use the Chi-square distribution for estimating standard deviation?
4. Can we construct a confidence interval for the mean as well?
5. What assumptions are needed to construct this confidence interval?

Tip: When calculating confidence intervals, always ensure that the data meets the assumptions of normality, as the Chi-square distribution is sensitive to this.