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. (round to two decimal places as needed.) a. the critical value is $z_0 = 1.64$. b. the critical values are $z_0 = pm square$. 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. $z < square$ b. $z > 1.64$ c. $z < square$ or $z > square$ (c) find the standardized test statistic $z$. $z = square$ (round to two decimal places as needed.)

Explanation:

Step1: Identify the test type

This is a two - sample z - test for the difference in means. The claim is that the precipitation in Seattle ($\mu_1$) is greater than in Birmingham ($\mu_2$), so the null hypothesis $H_0:\mu_1\leq\mu_2$ and the alternative hypothesis $H_a:\mu_1 > \mu_2$. We need the sample means of precipitation for Seattle ($\bar{x}_1$) and Birmingham ($\bar{x}_2$) from the data (not shown here, but let's assume we have them). The formula for the z - test statistic for two - sample means (when population standard deviations $\sigma_1$ and $\sigma_2$ are known) is:

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

Since $H_0$ assumes $\mu_1-\mu_2 = 0$, the formula simplifies to:

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

We know that $n_1 = n_2=30$, $\sigma_1 = 0.247$, $\sigma_2 = 0.523$. Let's assume we calculated $\bar{x}_1$ and $\bar{x}_2$ from the data. For example, if we assume (hypothetically, since data is not fully provided) $\bar{x}_1 = 0.15$ and $\bar{x}_2=0.10$ (these are just placeholders, in reality, we need to use the actual data). Then:

Step2: Plug in the values

First, calculate the denominator:

$$\sqrt{\frac{0.247^2}{30}+\frac{0.523^2}{30}}=\sqrt{\frac{0.061009 + 0.273529}{30}}=\sqrt{\frac{0.334538}{30}}=\sqrt{0.011151}\approx0.1056$$

Then the numerator: $\bar{x}_1-\bar{x}_2=0.15 - 0.10 = 0.05$ (again, placeholders). Then $z=\frac{0.05}{0.1056}\approx0.47$ (but this is with placeholder means). In reality, we need to use the actual sample means from the precipitation data. Let's assume the actual sample means are $\bar{x}_1 = 0.12$ and $\bar{x}_2 = 0.08$ (another placeholder). Then numerator is $0.12 - 0.08=0.04$. Denominator is still approximately $0.1056$. Then $z=\frac{0.04}{0.1056}\approx0.38$. But since the data is not fully provided, let's assume that after calculating with the actual data, the z - statistic is, for example, 1.80 (just for illustration, but in a real problem, we need to use the actual data).

(Note: Since the data is not fully shown, we can't calculate the exact value here, but the process is as above. Let's assume that after getting the sample means from the data, we plug into the formula. For the purpose of this example, let's say the correct calculation gives $z = 1.80$ (rounded to two decimal places).)

Answer:

(The actual answer depends on the sample means from the data. But following the formula, if we assume the correct calculation gives, for example, $z = 1.80$ (this is just an example, the real value requires the actual data), the answer would be $\boldsymbol{1.80}$ (or the correct value from the data - driven calculation).)