Hypothesis Test: Comparing IQ Scores by Blood Lead Levels Using Independent t-Test

Math Problem Statement

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). . . . Question content area top right Part 1 mu n x overbar s

Low Lead Level mu 1 95 92.82847 15.36311 High Lead Level mu 2 26 87.26269 8.62039 Question content area bottom Part 1 a. Use a 0.05 significance level to test the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels. What are the null and alternative hypotheses? Assume that population 1 consists of subjects with low lead levels and population 2 consists of subjects with high lead levels. A. Upper H 0: mu 1not equalsmu 2 Upper H 1: mu 1greater thanmu 2 B. Upper H 0: mu 1equalsmu 2 Upper H 1: mu 1not equalsmu 2 C. Upper H 0: mu 1equalsmu 2 Upper H 1: mu 1greater thanmu 2 Your answer is correct.D. Upper H 0: mu 1less than or equalsmu 2 Upper H 1: mu 1greater thanmu 2 Part 2 The test statistic is 2.41. (Round to two decimal places as needed.) Part 3 The P-value is . 009. (Round to three decimal places as needed.) Part 4 State the conclusion for the test. A. Reject the null hypothesis. There is sufficient evidence to support the claim that subjects with low lead levels have higher IQ scores. B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with low lead levels have higher IQ scores. C. Reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with low lead levels have higher IQ scores. D. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that subjects with low lead levels have higher IQ scores.

Solution

Let's go through each part step-by-step.

Part 1

We need to identify the correct null and alternative hypotheses.

Since we're testing if the mean IQ score of people with low blood lead levels is higher than that of people with high blood lead levels:

Population 1: Subjects with low lead levels ( $\mu_1$ )
Population 2: Subjects with high lead levels ( $\mu_2$ )

The hypotheses are:

Null Hypothesis ( $H_0$ ): $\mu_1 = \mu_2$
Alternative Hypothesis ( $H_1$ ): $\mu_1 > \mu_2$

So, the correct answer is:

C. $H_0: \mu_1 = \mu_2$ ; $H_1: \mu_1 > \mu_2$

Part 2

The test statistic $t$ has been calculated as $2.41$ .

This test statistic measures how many standard errors the sample mean difference is from zero (if the null hypothesis were true).

Part 3

The $p$ -value is given as $0.009$ , rounded to three decimal places.

This $p$ -value is less than the significance level ( $\alpha = 0.05$ ), suggesting that the result is statistically significant.

Part 4

Given the $p$ -value of $0.009$ , which is less than $0.05$ , we can reject the null hypothesis. Therefore, there is enough evidence to support the claim that subjects with low lead levels have higher IQ scores than those with high lead levels.

The correct conclusion is:

A. Reject the null hypothesis. There is sufficient evidence to support the claim that subjects with low lead levels have higher IQ scores.

Would you like more details on any part of this problem? Here are a few additional questions related to this analysis:

How would the test change if we were comparing if the IQ scores were simply different, not higher or lower?
What would happen if we assumed equal population standard deviations?
What does the $p$ -value represent in terms of probability for this context?
How would the results differ with a 0.01 significance level instead of 0.05?
Why is it important that the samples are independent?

Tip: When conducting hypothesis tests, always check the assumptions to ensure the test is valid, especially independence and normality in the samples.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Independent Samples t-Test
Significance Level
p-Value

Formulas

t = (x̄1 - x̄2) / sqrt((s1^2 / n1) + (s2^2 / n2))
p-value calculation for one-tailed t-test

Theorems

Central Limit Theorem
Student's t-Distribution

Suitable Grade Level

College

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