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. what is the rejection region? select the correct choice below and fill in the answer box(es) within your choice. (round to two decimal places as needed.) a. z < b. z > 1.64 c. z < or z > (c) find the standardized test statistic z. z = -1.02 (round to two decimal places as needed.) (d) decide whether to reject or fail to reject the null hypothesis. the p - value is (round to three decimal places as needed.)

Explanation:

Step1: Identify Test Type

This is a right - tailed z - test (since we're testing if precipitation in Seattle is greater than in Birmingham, so alternative hypothesis \(H_1:\mu_1>\mu_2\)). For a right - tailed test with \(\alpha = 0.05\), the critical value \(z_{\alpha}=1.645\) (approx 1.64). The standardized test statistic formula for two - sample z - test (when population standard deviations \(\sigma_1\) and \(\sigma_2\) are known) is \(z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\). Here, \(\mu_1-\mu_2 = 0\) (under null hypothesis \(H_0:\mu_1\leq\mu_2\)). But we need the P - value. For a z - statistic of \(z=- 1.02\) (wait, but if it's a right - tailed test and \(z=-1.02\), we need to find \(P(Z > - 1.02)\) or correct? Wait, no, the test statistic formula: let's assume \(\bar{x}_1\) is Seattle's mean, \(\bar{x}_2\) is Birmingham's mean. The formula is \(z=\frac{(\bar{x}_1-\bar{x}_2)-0}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\). But since we have \(z=-1.02\), for a right - tailed test, the P - value is \(P(Z > - 1.02)\). We know that \(P(Z > - 1.02)=1 - P(Z\leq - 1.02)\). Looking up in z - table, \(P(Z\leq - 1.02)=0.1539\), so \(P - value=1 - 0.1539 = 0.8461\)? Wait, no, wait, maybe I mixed up the direction. Wait, the alternative hypothesis is \(H_1:\mu_1>\mu_2\), so the test statistic is \(z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\). If \(z=-1.02\), that means \(\bar{x}_1-\bar{x}_2\) is negative, so the sample mean of Seattle is less than that of Birmingham. So for the right - tailed test, the P - value is \(P(Z > - 1.02)\). Let's calculate it properly. The z - score is \(z=-1.02\). The cumulative probability for \(z=-1.02\) is \(P(Z\leq - 1.02)\). From z - table, \(z=-1.02\): the row is - 1.0, column 0.02. The value is 0.1539. So \(P(Z > - 1.02)=1 - 0.1539 = 0.846\) (rounded to three decimal places).

Step2: Calculate P - value

Using the z - table or standard normal distribution properties. For \(z=-1.02\), \(P(Z\leq - 1.02)=0.1539\). Since the test is right - tailed (\(H_1:\mu_1>\mu_2\)), \(P - value=P(Z > z)=P(Z > - 1.02)=1 - P(Z\leq - 1.02)=1 - 0.1539 = 0.846\).

Answer:

0.846