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 σ₁ = $6521 and σ₂ = $6091. at α = 0.01, what should you conclude?

entry level architects in denver, co
x̄₁ = 58,300
n₁ = 33

entry level architects in lincoln, ne
x̄₂ = 54,240
n₂ = 35

what are the null and alternative hypotheses for this test?

a. h₀: μ₁ - μ₂ < 10,500; hₐ: μ₁ - μ₂ ≥ 10,500
b. h₀: μ₁ - μ₂ ≤ 10,500; hₐ: μ₁ - μ₂ > 10,500
c. h₀: μ₁ - μ₂ = 10,500; hₐ: μ₁ - μ₂ ≠ 10,500
d. h₀: μ₁ - μ₂ > 10,500; hₐ: μ₁ - μ₂ ≤ 10,500
e. h₀: μ₁ - μ₂ ≠ 10,500; hₐ: μ₁ - μ₂ = 10,500
f. h₀: μ₁ - μ₂ ≥ 10,500; hₐ: μ₁ - μ₂ < 10,500

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

Explanation:

Step1: Recall the formula for the z - statistic in two - sample z - test for the difference in means

The formula for the standardized test statistic (z - statistic) for testing the difference between two population means when the population standard deviations \(\sigma_1\) and \(\sigma_2\) are known is:

$$z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$$

In the null hypothesis \(H_0:\mu_1-\mu_2 = 10500\), so \((\mu_1 - \mu_2)=10500\).

We are given:
\(\bar{x}_1 = 58300\), \(\bar{x}_2=54240\), \(\sigma_1 = 6521\), \(\sigma_2 = 6091\), \(n_1 = 33\), \(n_2=35\)

Step2: Calculate the numerator

First, calculate \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\)

\(\bar{x}_1-\bar{x}_2=58300 - 54240=4060\)

\((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=4060-10500=- 6440\)

Step3: Calculate the denominator

Calculate \(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}\)

First, calculate \(\frac{\sigma_1^{2}}{n_1}=\frac{6521^{2}}{33}=\frac{42523441}{33}\approx1288589.12\)

\(\frac{\sigma_2^{2}}{n_2}=\frac{6091^{2}}{35}=\frac{37090281}{35}\approx1059722.31\)

Then, \(\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}=1288589.12 + 1059722.31=2348311.43\)

\(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{2348311.43}\approx1532.42\)

Step4: Calculate the z - statistic

Now, substitute the values of the numerator and the denominator into the z - formula:

\(z=\frac{-6440}{1532.42}\approx - 4.20\)

Answer:

\(z\approx - 4.20\)