Question

the annual salaries (in dollars) of 14 randomly chosen fire fighters are listed. at α = 0.05, is there enough evidence to support the claim that the standard deviation of the annual salaries is different from $5350? assume the population is normally distributed. complete parts (a) through (e) below.
50,762 40,962 52,386 46,539 41,734 40,173 51,138
52,013 43,827 34,908 35,096 28,222 32,693 37,832
click the icon to view the chi - square distribution table.
$\chi_{0}^{2} = 5.009, 24.736$
(round to three decimal places as needed. use a comma to separate answers as needed.)
identify the rejection region(s). choose the correct graph below.
a. b. c. d.
(c) find the standardized test statistic for the $\chi^{2}$ - test.
$\chi^{2} = \square$
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

We sum all 14 salaries and divide by 14 to get the sample mean \(\bar{x} \approx 42020.357\).

Step2: Calculate the sample variance

For each salary, we find the squared deviation from the mean, sum these squared deviations, and divide by \(n - 1 = 13\) to get \(s^2 \approx 60662440.77\).

Step3: Identify \(\sigma\) and \(n\)

We know \(\sigma = 5350\) (so \(\sigma^2 = 28622500\)) and \(n = 14\).

Step4: Calculate the \(\chi^2\)-test statistic

Using the formula \(\chi^2 = \frac{(n - 1)s^2}{\sigma^2}\), we substitute the values and compute the test statistic, which is approximately 27.55.

Answer:

To find the standardized test statistic for the \(\chi^2\)-test, we follow these steps:

Step 1: Calculate the sample variance (\(s^2\))

First, we need to find the sample mean (\(\bar{x}\)) of the 14 salaries:

$$ \bar{x} = \frac{50762 + 40962 + 52386 + 46539 + 41734 + 40173 + 51138 + 52013 + 43827 + 34908 + 35096 + 28222 + 32693 + 37832}{14} $$

Let's compute the sum:

$$
LATEXBLOCK0
$$

Now, the mean:

$$ \bar{x} = \frac{588285}{14} \approx 42020.357 $$

Next, we calculate the sample variance \(s^2\). The formula for sample variance is:

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

Let's compute each \((x_i - \bar{x})^2\):

1. \( (50762 - 42020.357)^2 \approx (8741.643)^2 \approx 76416320.0 \)
2. \( (40962 - 42020.357)^2 \approx (-1058.357)^2 \approx 1120120.0 \)
3. \( (52386 - 42020.357)^2 \approx (10365.643)^2 \approx 107446500.0 \)
4. \( (46539 - 42020.357)^2 \approx (4518.643)^2 \approx 20418100.0 \)
5. \( (41734 - 42020.357)^2 \approx (-286.357)^2 \approx 81990.0 \)
6. \( (40173 - 42020.357)^2 \approx (-1847.357)^2 \approx 3400700.0 \)
7. \( (51138 - 42020.357)^2 \approx (9117.643)^2 \approx 83131400.0 \)
8. \( (52013 - 42020.357)^2 \approx (9992.643)^2 \approx 99853000.0 \)
9. \( (43827 - 42020.357)^2 \approx (1806.643)^2 \approx 3263900.0 \)
10. \( (34908 - 42020.357)^2 \approx (-7112.357)^2 \approx 50585600.0 \)
11. \( (35096 - 42020.357)^2 \approx (-6924.357)^2 \approx 47946600.0 \)
12. \( (28222 - 42020.357)^2 \approx (-13798.357)^2 \approx 190405000.0 \)
13. \( (32693 - 42020.357)^2 \approx (-9327.357)^2 \approx 87000000.0 \)
14. \( (37832 - 42020.357)^2 \approx (-4188.357)^2 \approx 17542500.0 \)

Now, sum these squared deviations:

$$
LATEXBLOCK1
$$

Now, divide by \(n - 1 = 13\) to get the sample variance:

$$ s^2 = \frac{788611730}{13} \approx 60662440.77 $$

Step 2: Identify the hypothesized standard deviation (\(\sigma\)) and sample size (\(n\))

The claim is about the standard deviation being different from \(\$5350\), so \(\sigma = 5350\). The sample size \(n = 14\).

Step 3: Calculate the test statistic for the \(\chi^2\)-test

The formula for the \(\chi^2\)-test statistic for a test of variance (or standard deviation) is:

$$ \chi^2 = \frac{(n - 1)s^2}{\sigma^2} $$

First, compute \(\sigma^2\):

$$ \sigma^2 = (5350)^2 = 28622500 $$

Now, plug in the values:

$$ \chi^2 = \frac{(14 - 1) \times 60662440.77}{28622500} $$

Calculate the numerator:

$$ 13 \times 60662440.77 \approx 788611730 $$

Now, divide by \(\sigma^2\):

$$ \chi^2 = \frac{788611730}{28622500} \approx 27.55 $$

So, the standardized test statistic \(\chi^2 \approx 27.55\) (rounded to three decimal places).