Question

IQ and Lead Exposure Data Set 7 “IQ and Lead” in Appendix B lists full 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. The statistics are summarized on the top of the next page. Use a 0.05 significance level to test the claim that IQ scores of people with low lead

levels vary more than IQ scores of people with high lead levels.

Low Lead Level: n = 78, \(\bar x\) = 92.88462, s = 15.34451

High Lead Level: n = 21, \(\bar x\) = 86.90476, s = 8.988352

Short answer

There is enough evidence to support the claim that IQ scores of people with low lead levels vary more than IQ scores of people with high lead levels.

Step-by-step solution

Given information

For a sample of 78 IQ scores with low blood lead levels, the mean value equals92.88462,and the standard deviation equals15.34451. In another sample of 21 IQ scores with high blood lead levels, the mean valueequals86.90476,and the standard deviationequals8.988352.

It is claimed that the variation in the IQ scores for low blood lead levels is more than the variation in the IQ scores for high blood lead levels.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviationsof the IQ scores for low blood lead levels high blood lead levels,respectively.

Nullhypothesis:The population standard deviation of the IQ scores for low blood lead levels is more than the variation in the IQ scores for high blood lead levels.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

State the alternate hypotheses.

Since the original claim does not include equality, the alternate hypothesis\({H_1}\)represents the population variance of the low lead level is greater than the population variance of the high lead level.

Symbolically,

\({H_1}:{\sigma _1} > {\sigma _2}\)

Compute the test statistic for the t-test.

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {15.34451} \right)^2} = 235.454\)

\({\left( {8.988352} \right)^2} = 80.790\)

Here,\(s_1^2\)is the sample variance corresponding to low blood lead levels and has a value equal to 235.454.

\(s_2^2\)is the sample variance corresponding to high blood lead levels and has a value equal to 80.790.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{{{\left( {15.34451} \right)}^2}}}{{{{\left( {8.988352} \right)}^2}}}\\ = 2.914\end{array}\)

Thus, F is equal to 2.914.

Critical value and p-value

The value of the numerator degrees of freedom is equal tothe following:

\(\begin{array}{c}{n_1} - 1 = 78 - 1\\ = 77\end{array}\)

The value of the denominator degrees of freedom is equal tothe following:

\(\begin{array}{c}{n_2} - 1 = 21 - 1\\ = 20\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 77 and denominator degrees of freedom equal to 20 for a right-tailed test.

The level of significance is equal to 0.05.

Thus, the critical value is equal to 1.9246.

The two-tailed p-value for F equal to 2.914 is equal to 0.0045.

Conclusion

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to supportthe claimthat IQ scores of people with low lead levels vary more than IQ scores of people with high lead levels.