Question

a climatologist claims that the precipitation in seattle, washington, was greater than in birmingham, alabama, in a recent year. the daily precipitation amounts (in inches) for 30 days in a recent year in seattle and a recent year in birmingham are given in the accompanying table. assume the population standard deviation is 0.247 inch for seattle and 0.523 inch for birmingham. at α = 0.05, can you support the climatologist’s claim? complete parts (a) through (e). click the icon to view the precipitation data.

○ b. the mean precipitation in seattle was less than in birmingham in a recent year.
○ c. the mean precipitation in seattle was the same as in birmingham in a recent year.
○ d. the mean precipitation in seattle was different than in birmingham in a recent year.

let the data from seattle be sample 1 and let the data from birmingham be sample 2. identify h₀ and hₐ.

h₀: μ₁ ≤ μ₂
hₐ: μ₁ > μ₂

(b) find the critical value(s) and identify the rejection region. select the correct choice below and fill in the answer box within your choice.
(round to two decimal places as needed.)

● a. the critical value is z₀ = .
○ b. the critical values are z₀ = ± .

Explanation:

Step1: Identify Test Type

This is a right - tailed z - test (since population standard deviations are known, sample size \(n = 30\) for both, and we are testing if \(\mu_1>\mu_2\)). For a right - tailed test with \(\alpha=0.05\), we need to find the critical value \(z_0\) such that \(P(Z > z_0)=\alpha = 0.05\).

Step2: Use Standard Normal Table

The cumulative probability for the critical value in a right - tailed test with \(\alpha = 0.05\) is \(1-\alpha=0.95\). Looking up the \(z\) - value in the standard normal distribution table (or using a calculator with a normal - distribution function) that corresponds to a cumulative probability of \(0.95\), we find that \(z_0 = 1.645\) (because \(P(Z\leq1.645)=0.95\), so \(P(Z > 1.645)=0.05\)).

Answer:

The critical value is \(z_0=\boxed{1.645}\)