Question

is the difference between the mean annual salaries of entry level architects in denver, colorado, and lincoln, nebraska, equal to $10,500? to decide, you select a random sample of entry level architects from each city. the results of each survey are shown. assume the population standard deviations are $\sigma_1 = \\$6521$ and $\sigma_2 = \\$6091$. at $\alpha = 0.01$, what should you conclude?

entry level architects in denver, co
$\bar{x}_1 = 58,300$
$n_1 = 33$

entry level architects in lincoln, ne
$\bar{x}_2 = 54,240$
$n_2 = 35$

\bigcirc d. $h_0: \mu_1 - \mu_2 > 10,500$
$h_a: \mu_1 - \mu_2 \leq 10,500$
\bigcirc e. $h_0: \mu_1 - \mu_2 \
eq 10,500$
$h_a: \mu_1 - \mu_2 = 10,500$
\bigcirc f. $h_0: \mu_1 - \mu_2 \geq 10,500$
$h_a: \mu_1 - \mu_2 < 10,500$

calculate the standardized test statistic.
$z = -4.20$
(round to two decimal places as needed.)

determine the p - value.
the p - value is \square.
(round to three decimal places as needed.)

Explanation:

Step1: Identify Test Type

This is a two - sample z - test for the difference between two means. The standardized test statistic \(z=- 4.20\) is given. For a two - tailed test (since we are testing if the difference is not equal to 10500, the alternative hypothesis \(H_{a}:\mu_{1}-\mu_{2}
eq10500\)), we need to find the area in both tails corresponding to \(z = - 4.20\) (and \(z = 4.20\) because of symmetry).

Step2: Use Z - table or Calculator

We know that the standard normal distribution is symmetric. The P - value for a two - tailed test is \(2\times P(Z < -|z|)\). For \(z=-4.20\), we look up the probability that \(Z < - 4.20\) in the standard normal table. The probability that \(Z < - 4.20\) is extremely small. Using a calculator or a more detailed z - table, we find that \(P(Z < - 4.20)\approx0.000013\). Then the P - value for a two - tailed test is \(2\times0.000013 = 0.000026\approx0.000\) (rounded to three decimal places).

Answer:

\(0.000\)