In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

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 below.

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.

b. Construct a confidence interval appropriate for the hypothesis test in part (a).

c. Does exposure to lead appear to have an effect on IQ scores?

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

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

For a sample of 78 IQ scores with low blood lead levels, the mean value is equal to 92.88462, and the standard deviation is equal to 15.34451. In another sample of 21 IQ scores with high blood lead levels, the mean value is equal to 86.90476, and the standard deviation is equal to 8.988352.

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

Null hypothesis:The population mean of the IQ scores for low blood lead levels is equal to the population mean of the IQ scores for high blood lead levels.

Symbolically,

\({H_0}:{\mu _1} = {\mu _2}\).

Alternative hypothesis: The population mean of the IQ scores for low blood lead levels is greater than the population mean of the IQ scores for high blood lead levels.

Symbolically,

\({H_1}:{\mu _1} > {\mu _2}\).

As the alternative hypothesis contains the greater-than symbol, it is aright-tailed test.

The sample mean corresponding to the low blood lead levels\(\left( {{{\bar x}_1}} \right)\)is equal to 92.8862.

The sample mean corresponding to the high blood lead levels\(\left( {{{\bar x}_2}} \right)\)is equal to 86.90476.

The sample standard deviation\(\left( {{s_1}} \right)\)corresponding to the low blood lead levels is equal to 15.34451.

The sample standard deviation\(\left( {{s_2}} \right)\)corresponding to the high blood lead levels is equal to 8.988352.

The sample size\(\left( {{n_1}} \right)\)is equal to 78, and the sample size\(\left( {{n_2}} \right)\)is equal to 21.

The test statistic is as follows.

Substitute the respective values in the above formula and simplify.

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}}} }}\\ = \frac{{\left( {92.88462 - 86.90476} \right) - 0}}{{\sqrt {\frac{{{{15.34451}^2}}}{{78}} + \frac{{{{8.988352}^2}}}{{21}}} }}\\ = 2.282\end{array}\).

Thus, t is equal to 2.282.

The degree of freedom is the smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below.

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

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

Thus, the value of the degrees of freedom is equal to the smaller one of the values 77 and 20, which is 20.

The critical value can be obtained using the t-distribution table with degrees of freedom equal to 20 and the significance level equal to 0.05 for a right-tailed test.

Thus, the critical value is equal to 1.7247.

The p-value for t equal to 2.828 is equal to 0.0168.

a.

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

Therefore,there is sufficient evidence to support 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.

b.

The confidence interval has the following expression:

\(CI = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\).

If a one-tailed hypothesis test is conducted at a 0.05 level of significance, the confidence level to construct the confidence interval is equal to 90%.

Thus, the level of significance to construct the confidence interval becomes\(\alpha = 0.10\).

The margin of error is given by the following formula:
\(E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \).

Substitute the respective values in the above formula to compute the margin of error.

\(\begin{array}{c}E = 1.7247 \times \sqrt {\frac{{{{15.34451}^2}}}{{78}} + \frac{{{{8.988352}^2}}}{{21}}} \\ = 4.51918\end{array}\).

Substituting the value of E and the sample means, the following confidence interval is obtained:

\(\begin{array}{c}\left( {92.88462 - 86.90476} \right) - 4.51918 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {92.88462 - 86.90476} \right) + 4.51918\\1.46 < \left( {{\mu _1} - {\mu _2}} \right) < 10.50\end{array}\)

Thus, the 90% confidence interval of the difference in the two population means is equal to (1.46, 10.50).

It can be observed that the interval does not contain 0.

Thus, the mean IQ score of people with low blood lead levels cannot be equal to the mean IQ score of people with high blood lead levels.

Therefore,there is sufficient evidence to support 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.

c.

People with low blood lead levels have lower mean IQ scores than people with high blood lead levels.

Therefore, blood lead levels have a significant effect on IQ scores.