Question

a politician claims that the mean salary for managers in his state is more than the national mean, $83,000. assume the the population is normally distributed and the population standard deviation is $7700. the salaries (in dollars) for a random sample of 30 managers in the state are listed. at \\( \alpha = 0.07 \\), is there enough evidence to support the claim? use technology.

(a) identify the null hypothesis and alternative hypothesis.

a. \\( h _ { 0 } : \mu \leq 83,000 \\)
\\( h _ { a } : \mu > 83,000 \\)

b. \\( h _ { 0 } : \mu \
eq 83,000 \\)
\\( h _ { a } : \mu = 83,000 \\)

c. \\( h _ { 0 } : \mu > 83,000 \\)
\\( h _ { a } : \mu \leq 83,000 \\)

d. \\( h _ { 0 } : \mu = 83,000 \\)
\\( h _ { a } : \mu \
eq 83,000 \\)

e. \\( h _ { 0 } : \mu > 83,000 \\)
\\( h _ { a } : \mu \leq 83,000 \\)

f. \\( h _ { 0 } : \mu \geq 83,000 \\)
\\( h _ { a } : \mu < 83,000 \\)

(b) identify the standardized test statistic.

\\( z = \\)
(round the final answer to two places as needed. round all intermediate values to three places as needed.)

Explanation:

Step1: Calculate the sample mean

Using technology (e.g., a calculator or statistical software), input the given data values. Let \(x_1,x_2,\cdots,x_{30}\) be the salary values. The formula for the sample mean \(\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i\), where \(n = 30\).

Step2: Calculate the standardized test statistic \(z\)

The formula for the \(z\) - statistic in a one - sample z - test (since the population standard deviation \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\mu = 83000\), \(\sigma=7700\), and \(n = 30\).

Using a calculator or statistical software to calculate the sample mean \(\bar{x}\) from the data:
\(\bar{x}=\frac{78945 + 83509+89466+79248+98556+80394+76746+75022+88173+81659+77026+87817+83225+97502+85025+82836+81053+91161+71290+81179+72479+86600+81846+75618+89265+70529+88004+87083+78496+98930}{30}\)
\(\bar{x}=\frac{2477919}{30}=82597.3\)

Then, \(z=\frac{82597.3 - 83000}{\frac{7700}{\sqrt{30}}}\)
First, calculate \(\frac{7700}{\sqrt{30}}\approx\frac{7700}{5.477}\approx1406.06\)
Then \(z=\frac{82597.3 - 83000}{1406.06}=\frac{- 402.7}{1406.06}\approx - 0.29\)

Answer:

\(z\approx - 0.29\)