Researchers claim that the birth rate in Bonn, Germany is higher than the national average. A random sample of 1400 Bonn residents had 14 births, whereas a random sample of 4000 people from all over Germany had 24 births during the same year. Test the researchers’ claim using a 0.05 level of significance. Let Bonn residents be Population 1 and let people from all over Germany be Population 2.
Step 2 of 3 :  Compute the value of the test statistic. Round your answer to two decimal places.

To test the claim that the birth rate in Bonn (Population 1) is higher than the national average (Population 2) at a \(0.05\) level of significance, we can set up a hypothesis test for two population proportions. Let's follow the steps to calculate the test statistic.

### Step 1: Define Hypotheses
Let:
- \( p_1 \) = proportion of births in Bonn
- \( p_2 \) = proportion of births in Germany

Our hypotheses are:
- Null Hypothesis: \( H_0: p_1 \leq p_2 \)
- Alternative Hypothesis: \( H_1: p_1 > p_2 \)

### Step 2: Calculate Sample Proportions
Given:
- For Population 1 (Bonn): \( x_1 = 14 \) births, \( n_1 = 1400 \)
- For Population 2 (Germany): \( x_2 = 24 \) births, \( n_2 = 4000 \)

The sample proportions are:
\[
\hat{p}_1 = \frac{x_1}{n_1} = \frac{14}{1400} = 0.01
\]
\[
\hat{p}_2 = \frac{x_2}{n_2} = \frac{24}{4000} = 0.006
\]

### Step 3: Calculate the Pooled Proportion
Since we assume the null hypothesis is true for the calculation, we use a pooled proportion \( \hat{p} \):
\[
\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{14 + 24}{1400 + 4000} = \frac{38}{5400} \approx 0.00704
\]

### Step 4: Compute the Standard Error
The standard error (SE) of the difference between the two proportions is:
\[
SE = \sqrt{\hat{p} \left(1 - \hat{p}\right) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]
Substitute the values:
\[
SE = \sqrt{0.00704 \times (1 - 0.00704) \times \left(\frac{1}{1400} + \frac{1}{4000}\right)}
\]
\[
SE = \sqrt{0.00704 \times 0.99296 \times (0.0007143 + 0.00025)}
\]
\[
SE \approx \sqrt{0.00704 \times 0.99296 \times 0.0009643} \approx \sqrt{0.00000675} \approx 0.0026
\]

### Step 5: Calculate the Test Statistic
The test statistic \( Z \) for comparing two proportions is given by:
\[
Z = \frac{\hat{p}_1 - \hat{p}_2}{SE}
\]
Substitute the values:
\[
Z = \frac{0.01 - 0.006}{0.0026} \approx \frac{0.004}{0.0026} \approx 1.54
\]

### Answer
The value of the test statistic is approximately:
\[
Z \approx 1.54
\]

Would you like me to continue with interpretation or calculations for the p-value?

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#### Related Questions
1. What would be the next steps to determine if we reject the null hypothesis?
2. How would changing the significance level affect our decision?
3. Can this method be applied if the sample sizes were different?
4. What are the assumptions underlying this test?
5. How does the sample size impact the test statistic's sensitivity?

#### Tip
To strengthen your understanding, try solving a similar problem with different sample sizes or proportions.

Researchers claim that the birth rate in Bonn, Germany is higher than the national average. A random sample of 1400 Bonn residents had 14 births, whereas a random sample of 4000 people from all over Germany had 24 births during the same year. Test the researchers’ claim using a 0.05 level of significance. Let Bonn residents be Population 1 and let people from all over Germany be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.

Conduct a hypothesis test to evaluate if the birth rate in Bonn, Germany exceeds the national average using sample data. A detailed step-by-step solution covers calculating sample proportions, pooled proportion, standard error, and Z-test statistic. Suitable for college-level statistics.

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