Question

the annual salaries (in dollars) of 14 randomly chosen fire fighters are listed. at α = 0.05, is there enough evidence support the claim that the standard deviation of the annual salaries is different from $5350? assume the populatior 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 32,683 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. \\(\bigcirc\\) a. \\(\bigcirc\\) b. \\(\bigcirc\\) c. \\(\bigcirc\\) d. (c) find the standardized test statistic for the \\(chi^{2}\\)-test. \\(chi^{2} = \square\\) (round to three decimal places as needed.)

Explanation:

Answer:

To find the standardized test statistic \(\chi^{2}\) for the test of the standard deviation, we use the formula:

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

where:

• \(n = 14\) (sample size),
• \(s^{2}\) is the sample variance,
• \(\sigma^{2}=(5350)^{2}\) (claimed population variance, since the claimed standard deviation \(\sigma = 5350\)).

Step 1: Calculate the sample mean \(\bar{x}\)

First, we find the sample mean of the given salaries:

$$ \bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n} $$

Summing the salaries:

$$
LATEXBLOCK0
$$

$$ \bar{x}=\frac{588275}{14}\approx42019.6429 $$

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

The sample variance is given by:

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

We calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\):

• For \(x_{1}=50762\): \((50762 - 42019.6429)^{2}=(8742.3571)^{2}\approx76428797.5\)
• For \(x_{2}=40962\): \((40962 - 42019.6429)^{2}=(- 1057.6429)^{2}\approx1118610.0\)
• For \(x_{3}=52386\): \((52386 - 42019.6429)^{2}=(10366.3571)^{2}\approx107461350.0\)
• For \(x_{4}=46539\): \((46539 - 42019.6429)^{2}=(4519.3571)^{2}\approx20424600.0\)
• For \(x_{5}=41734\): \((41734 - 42019.6429)^{2}=(-285.6429)^{2}\approx81592.0\)
• For \(x_{6}=40173\): \((40173 - 42019.6429)^{2}=(-1846.6429)^{2}\approx3409900.0\)
• For \(x_{7}=51138\): \((51138 - 42019.6429)^{2}=(9118.3571)^{2}\approx83144400.0\)
• For \(x_{8}=52013\): \((52013 - 42019.6429)^{2}=(9993.3571)^{2}\approx99867000.0\)
• For \(x_{9}=43827\): \((43827 - 42019.6429)^{2}=(1807.3571)^{2}\approx3266500.0\)
• For \(x_{10}=34908\): \((34908 - 42019.6429)^{2}=(-7111.6429)^{2}\approx50575500.0\)
• For \(x_{11}=35096\): \((35096 - 42019.6429)^{2}=(-6923.6429)^{2}\approx47937000.0\)
• For \(x_{12}=28222\): \((28222 - 42019.6429)^{2}=(-13797.6429)^{2}\approx190375000.0\)
• For \(x_{13}=32683\): \((32683 - 42019.6429)^{2}=(-9336.6429)^{2}\approx87161000.0\)
• For \(x_{14}=37832\): \((37832 - 42019.6429)^{2}=(-4187.6429)^{2}\approx17536400.0\)

Summing these squared deviations:

$$
LATEXBLOCK1
$$

$$ s^{2}=\frac{788787649.5}{14 - 1}=\frac{788787649.5}{13}\approx60675973.04 $$

Step 3: Calculate the test statistic \(\chi^{2}\)

We know that \(\sigma = 5350\), so \(\sigma^{2}=(5350)^{2}=28622500\)

$$ \chi^{2}=\frac{(14 - 1)\times60675973.04}{28622500} $$

$$ \chi^{2}=\frac{13\times60675973.04}{28622500}=\frac{788787649.52}{28622500}\approx27.558 $$

The standardized test statistic \(\chi^{2}\approx\boldsymbol{27.558}\) (rounded to three decimal places)