To construct a [tex]$90\%$[/tex] confidence interval for the population standard deviation [tex]\(\sigma\)[/tex] at Bank A, we can follow these steps:
### Step 1: Calculate the Sample Standard Deviation
First, we need to find the sample standard deviation of the waiting times at Bank A. The sample waiting times are:
[tex]\[ 5.7, 8.9, 2.6, 9.0, 6.9, 9.6, 2.5, 9.2, 7.3, 12.0 \][/tex]
From these values, the sample standard deviation ([tex]\(s\)[/tex]) is:
[tex]\[ s = 3.06 \][/tex]
### Step 2: Determine the Degrees of Freedom
The degrees of freedom ([tex]\(df\)[/tex]) for the sample are calculated as the sample size minus one. If our sample size ([tex]\(n\)[/tex]) is 10, then:
[tex]\[ df = n - 1 = 10 - 1 = 9 \][/tex]
### Step 3: Find the Chi-Square Critical Values
For a [tex]$90\%$[/tex] confidence interval, we split the significance level ([tex]\(\alpha = 0.10\)[/tex]) into two tails, so [tex]\(\alpha/2 = 0.05\)[/tex]. We need the critical values for [tex]\(\chi^2\)[/tex] distribution with [tex]\(df = 9\)[/tex]:
- Lower critical value ([tex]\(\chi^2_{lower}\)[/tex]) at [tex]\(0.05\)[/tex] significance level:
[tex]\[ \chi^2_{0.05, 9} = 3.33 \][/tex]
- Upper critical value ([tex]\(\chi^2_{upper}\)[/tex]) at [tex]\(0.95\)[/tex] significance level:
[tex]\[ \chi^2_{0.95, 9} = 16.92 \][/tex]
### Step 4: Calculate the Confidence Interval for the Population Standard Deviation
Using the critical values and the sample standard deviation, we calculate the confidence interval for the population standard deviation ([tex]\(\sigma\)[/tex]) using the following formulas:
[tex]\[ \sigma_{lower} = \sqrt{\left(\frac{(n-1) \cdot s^2}{\chi^2_{upper}}\right)} \][/tex]
[tex]\[ \sigma_{upper} = \sqrt{\left(\frac{(n-1) \cdot s^2}{\chi^2_{lower}}\right)} \][/tex]
Given:
- [tex]\( s = 3.06 \)[/tex]
- [tex]\(\chi^2_{lower} = 3.33\)[/tex]
- [tex]\(\chi^2_{upper} = 16.92\)[/tex]
- [tex]\( df = 9\)[/tex]
### Step 5: Perform the Calculations
[tex]\[ \sigma_{lower} = \sqrt{\left(\frac{9 \cdot (3.06)^2}{16.92}\right)} = 2.23 \][/tex]
[tex]\[ \sigma_{upper} = \sqrt{\left(\frac{9 \cdot (3.06)^2}{3.33}\right)} = 5.03 \][/tex]
### Conclusion
Hence, the [tex]$90\%$[/tex] confidence interval for the population standard deviation [tex]\(\sigma\)[/tex] at Bank A is:
[tex]\[ 2.23 < \sigma < 5.03 \][/tex]
Rounded to two decimal places, the confidence interval is:
[tex]\[ 2.23 \text{ min} < \sigma < 5.03 \text{ min} \][/tex]
