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.
78,945 83,509 89,466 79,248 98,556 80,394 76,746 75,022 88,173 81,659
77,026 87,817 83,225 97,502 85,025 82,836 81,053 91,161 71,290 81,179
72,479 86,600 81,846 75,618 89,265 70,529 88,004 87,083 78,496 98,930
(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 = 0.21 \\)
(round the final answer to two places as needed. round all intermediate values to three places as needed.)
(c) find the p - value. use technology.
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for P - value in a right - tailed z - test

For a right - tailed z - test with test statistic \(z\), the P - value is \(P(Z>z)\), where \(Z\) is a standard normal random variable.

Step2: Use the standard normal distribution table or technology

We know \(z = 0.21\). Using technology (e.g., a TI - 84 Plus: normalcdf(0.21,1000) or Excel: =1 - NORM.S.DIST(0.21,TRUE)), we calculate the P - value.

$$P - value=1-\Phi(0.21)$$

where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.

$$1 - 0.583176=0.416824$$

Answer:

\(0.417\)