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Use concepts on hypothesis testing and sampling distribution for one and two means to solve and compute.

This is a hypothesis testing problem involving two independent samples. We are asked to determine if there is a significant difference in the mean blood lead levels between schoolchildren and street child vendors. Let me outline the steps and solve the problem.

### Given Data:
- **Group 1: Schoolchildren**
  - \( n_1 = 414 \)
  - \( \bar{X}_1 = 14.1 \, \mu g/dl \) (mean blood lead level)
  - \( s_1 = 5.3 \, \mu g/dl \) (standard deviation)

- **Group 2: Street child vendors**
  - \( n_2 = 101 \)
  - \( \bar{X}_2 = 15.0 \, \mu g/dl \) (mean blood lead level)
  - \( s_2 = 5.5 \, \mu g/dl \) (standard deviation)

- **Significance level**: \( \alpha = 0.01 \)

We will use a two-sample t-test assuming equal population variances.

### Hypotheses:
- Null Hypothesis \( H_0: \mu_1 = \mu_2 \) (There is no difference in mean blood lead levels between the two groups.)
- Alternative Hypothesis \( H_1: \mu_1 \neq \mu_2 \) (There is a difference in mean blood lead levels.)

### Step 1: Calculate the test statistic

For a two-sample t-test with equal variances, the formula for the test statistic is:

\[
t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]

Where:
- \( s_p \) is the pooled standard deviation, calculated as:

\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\]

### Step 2: Degrees of freedom

The degrees of freedom (df) for this test is:

\[
df = n_1 + n_2 - 2
\]

Now, let's compute these values.The computed values are as follows:
- The pooled standard deviation \( s_p = 5.34 \, \mu g/dl \)
- The test statistic \( t = -1.52 \)
- The degrees of freedom \( df = 513 \)

### Step 3: Determine the critical value or p-value

Using a two-tailed test with \( \alpha = 0.01 \) and degrees of freedom \( df = 513 \), we can find the critical t-value from the t-distribution table or calculate the p-value. I'll now compute the p-value for this test.The p-value for the test is approximately \( 0.129 \).

### Step 4: Conclusion
Since the p-value \( 0.129 \) is greater than the significance level \( \alpha = 0.01 \), we **fail to reject the null hypothesis**. This means that there is not enough evidence to conclude a significant difference in the mean blood lead levels between schoolchildren and street child vendors at the 1% significance level.

Would you like further details on any step or have any questions?

### Related Questions:
1. What is the impact of changing the significance level \( \alpha \) on the results?
2. How would the conclusion change if population variances were unequal?
3. What assumptions are made in a two-sample t-test with equal variances?
4. How is the pooled standard deviation used in hypothesis testing?
5. Can you apply a one-tailed test in this scenario?

### Tip:
When conducting hypothesis tests, always check the assumptions underlying the test (such as normality and equal variances) to ensure valid results.

One of the objectives of a recent study on environmental pollution is to assess the health effects of vehicular emissions on vulnerable population groups. The authors took interest in looking at the mean blood lead levels of two groups of children: schoolchildren and street child vendors. In a sample of 414 schoolchildren, the mean blood level was 14.1 μg/dl and a standard deviation of 5.3 μg/dl. Among 101 street child vendors, the mean blood lead level was 15.0 μg/dl with a standard deviation of 5.5 μg/dl. Assuming that the population variances are equal, do these data provide sufficient evidence to indicate a difference in the mean blood lead level in the population represented? Let α=0.01.

This statistical problem assesses the difference in mean blood lead levels between two groups: schoolchildren and street child vendors. Using a two-sample t-test with equal variances, we explore whether the observed differences are statistically significant, given the sample data and a significance level of α=0.01.

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