The IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood were collected. The statistics are summarized in the accompanying table. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts​ (a) to​ (c). 	μ	n	@OVERBAR{x}	s
Low Lead Level	μ@Sub{1}	83	94.20476	15.14932
High Lead Level	μ@Sub{2}	21	87.58119	9.83286The test statistic is 
  
2.44. ​(Round to two decimal places as​ needed.)
Part 3
The​ P-value is 
  
enter your response here. ​(Round to three decimal places as​ needed.)

To complete part (c), we need to calculate the \( p \)-value based on the given test statistic and details. Here’s how to approach it:

1. **Given Data**:
   - Sample 1 (Low Lead Level): 
     - \( n_1 = 83 \)
     - \( \overline{x}_1 = 94.20476 \)
     - \( s_1 = 15.14932 \)
   - Sample 2 (High Lead Level): 
     - \( n_2 = 21 \)
     - \( \overline{x}_2 = 87.58119 \)
     - \( s_2 = 9.83286 \)
   - Test statistic, \( t = 2.44 \)

2. **Determine the degrees of freedom**: 
   Since we are not assuming equal variances, we use the following approximation formula for the degrees of freedom (\( df \)):

   \[
   df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}
   \]

3. **Calculate the \( p \)-value**:
   Using the calculated degrees of freedom and the test statistic of \( t = 2.44 \), we can look up the \( p \)-value from the \( t \)-distribution table or use a calculator.

Let’s perform the calculations to find the \( p \)-value.The approximate degrees of freedom for this test are \( df \approx 47.09 \), and the \( p \)-value is approximately \( 0.019 \) (rounded to three decimal places).

Would you like additional details on these calculations or have any questions?

Here are some related questions for further exploration:
1. What does a \( p \)-value of 0.019 indicate in the context of hypothesis testing?
2. How do we interpret the degrees of freedom in a two-sample \( t \)-test with unequal variances?
3. What assumptions are necessary for conducting a two-sample \( t \)-test?
4. How would the test change if we assumed equal variances between the two groups?
5. What are the consequences of sample size differences in hypothesis testing?

**Tip:** For small samples, the assumption of normality is crucial in \( t \)-tests. Always examine this assumption to ensure valid results.

The IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood were collected. The statistics are summarized in the accompanying table. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts​ (a) to​ (c).

Explore a two-sample t-test to compare IQ scores between groups with low and high lead levels in their blood. This problem covers hypothesis testing concepts, including calculating degrees of freedom and interpreting the p-value for unequal variances. Suitable for students in introductory statistics.

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