Math / Data & Statistics

Question```latex 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 0.248 inch for Seattle and 0.519 inch for Birmingham. At $\alpha=0.05$ , can you support the climatologist's claim? Complete parts (a) through (e).
\begin{tabular}{|lllllllllllll|} \hline Seattle: & 0.003 & 0.002 & 0.054 & 0.010 & 0.211 & 0.001 & 0.003 & 0.518 & 0.003 & 0.014 \\ & 0.000 & 0.193 & 0.004 & 0.184 & 0.017 & 0.018 & 0.127 & 0.000 & 0.032 & 0.002 \\ & 0.037 & 0.003 & 0.412 & 0.227 & 0.003 & 0.804 & 0.150 & 0.002 & 0.003 & 0.790 \\ Birmingham: & 0.001 & 0.957 & 0.836 & 0.000 & 0.099 & 0.003 & 0.003 & 0.204 & 0.004 & 0.541 \\ & 0.967 & 0.001 & 0.351 & 0.020 & 0.043 & 0.697 & 0.003 & 0.004 & 0.000 & 0.000 \\ & 0.030 & 0.009 & 0.152 & 0.273 & 0.004 & 0.002 & 0.933 & 0.003 & 0.892 & 0.008 \\ \hline \end{tabular}
Click the icon to view the precipitation data.
A. The critical values are $z_{0}= \pm$ $\square$
B. The critical value is $z_{0}=1.65$
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. $z<$ $\square$ or $z>$ $\square$
B. $z<$ $\square$
C. $z>$ $\square$
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Studdy Solution

STEP 1

1. We are conducting a hypothesis test to compare the mean daily precipitation between Seattle and Birmingham.
2. The population standard deviations are known: $\sigma_{\text{Seattle}} = 0.248$ and $\sigma_{\text{Birmingham}} = 0.519$ .
3. The significance level is $\alpha = 0.05$ .
4. The sample size for both Seattle and Birmingham is 30 days.

STEP 2

1. State the null and alternative hypotheses.
2. Determine the critical value and rejection region.
3. Calculate the test statistic.
4. Make a decision to reject or not reject the null hypothesis.
5. Interpret the results.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean precipitation in Seattle is less than or equal to that in Birmingham. - Alternative hypothesis ( $H_a$ ): The mean precipitation in Seattle is greater than that in Birmingham.
$H_0: \mu_{\text{Seattle}} \leq \mu_{\text{Birmingham}}$ $H_a: \mu_{\text{Seattle}} > \mu_{\text{Birmingham}}$

STEP 4

Determine the critical value for a one-tailed test at $\alpha = 0.05$ .
- Since the test is one-tailed, we use the critical value $z_0 = 1.65$ .
The rejection region is:
$z > 1.65$

STEP 5

Calculate the test statistic using the formula for the difference of means with known population standard deviations:
$z = \frac{(\bar{x}_{\text{Seattle}} - \bar{x}_{\text{Birmingham}})}{\sqrt{\frac{\sigma_{\text{Seattle}}^2}{n} + \frac{\sigma_{\text{Birmingham}}^2}{n}}}$
Where: - $\bar{x}_{\text{Seattle}}$ is the sample mean for Seattle. - $\bar{x}_{\text{Birmingham}}$ is the sample mean for Birmingham. - $n = 30$ .
Calculate the sample means and substitute the values to find $z$ .

STEP 6

Calculate the sample means:
$\bar{x}_{\text{Seattle}} = \frac{\sum \text{Seattle data}}{30}$ $\bar{x}_{\text{Birmingham}} = \frac{\sum \text{Birmingham data}}{30}$
Substitute these means into the test statistic formula and calculate $z$ .

STEP 7

Compare the calculated $z$ value to the critical value $1.65$ .
- If $z > 1.65$ , reject the null hypothesis. - Otherwise, do not reject the null hypothesis.

STEP 8

Interpret the results:
- If the null hypothesis is rejected, conclude that there is sufficient evidence to support the climatologist's claim that precipitation in Seattle was greater than in Birmingham. - If the null hypothesis is not rejected, conclude that there is not enough evidence to support the claim.

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Studdy Solution

STEP 1

STEP 2

1. State the null and alternative hypotheses.
2. Determine the critical value and rejection region.
3. Calculate the test statistic.
4. Make a decision to reject or not reject the null hypothesis.
5. Interpret the results.

STEP 3

STEP 4

Determine the critical value for a one-tailed test at $\alpha = 0.05$ .
- Since the test is one-tailed, we use the critical value $z_0 = 1.65$ .
The rejection region is:
$z > 1.65$

STEP 5

STEP 6

Calculate the sample means:
$\bar{x}_{\text{Seattle}} = \frac{\sum \text{Seattle data}}{30}$ $\bar{x}_{\text{Birmingham}} = \frac{\sum \text{Birmingham data}}{30}$
Substitute these means into the test statistic formula and calculate $z$ .

STEP 7

Compare the calculated $z$ value to the critical value $1.65$ .
- If $z > 1.65$ , reject the null hypothesis. - Otherwise, do not reject the null hypothesis.

STEP 8

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Studdy Solution

STEP 1

STEP 2

1. State the null and alternative hypotheses.2. Determine the critical value and rejection region.3. Calculate the test statistic.4. Make a decision to reject or not reject the null hypothesis.5. Interpret the results.

STEP 3

STEP 4

Determine the critical value for a one-tailed test at α=0.05 \alpha = 0.05 α=0.05.- Since the test is one-tailed, we use the critical value z0=1.65 z_0 = 1.65 z0​=1.65.The rejection region is:z>1.65 z > 1.65 z>1.65

STEP 5

STEP 6

STEP 7

Compare the calculated z z z value to the critical value 1.65 1.65 1.65.- If z>1.65 z > 1.65 z>1.65, reject the null hypothesis. - Otherwise, do not reject the null hypothesis.

STEP 8

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. State the null and alternative hypotheses.
2. Determine the critical value and rejection region.
3. Calculate the test statistic.
4. Make a decision to reject or not reject the null hypothesis.
5. Interpret the results.

Determine the critical value for a one-tailed test at $\alpha = 0.05$ .
- Since the test is one-tailed, we use the critical value $z_0 = 1.65$ .
The rejection region is:
$z > 1.65$

Compare the calculated $z$ value to the critical value $1.65$ .
- If $z > 1.65$ , reject the null hypothesis. - Otherwise, do not reject the null hypothesis.