Question

Repeat Exercise 12 “IQ and Lead” by assuming that the two population standard deviations are equal, so \({\sigma _1} = {\sigma _2}\). Use the appropriate method from Part 2 of this section. Does pooling the standard deviations yield results showing greater significance?

Short answer

No, pooling the standard deviations does not show results with greater significance.

Step-by-step solution

Given information

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.

Hypothesis

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.

Compute the pooled sample variance

The pooled sample variance is given as follows.

\(\begin{array}{c}{s_p}^2 = \frac{{\left( {{n_1} - 1} \right)s_1^2 + \left( {{n_2} - 1} \right)s_2^2}}{{\left( {{n_1} - 1} \right) + \left( {{n_2} - 1} \right)}}\\ = \frac{{\left( {78 - 1} \right){{\left( {15.34451} \right)}^2} + \left( {21 - 1} \right){{\left( {8.988352} \right)}^2}}}{{\left( {78 - 1} \right) + \left( {21 - 1} \right)}}\\ = 203.56\end{array}\).

Thus, the pooled sample variance is equal to 203.56.

Compute the test statistic

The following values are obtained from exercise 12.

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 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 equal to

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_p^2}}{{{n_1}}} + \frac{{s_p^2}}{{{n_2}}}} }}\\ = \frac{{92.88462 - 86.90476}}{{\sqrt {\frac{{203.56}}{{78}} + \frac{{203.56}}{{21}}} }}\\ = 1.7048\end{array}\)

Thus, the test statistic is 1.7048.

Compute the degree of freedom

The degree of freedom is equal to

\(\begin{array}{c}df = {n_1} + {n_2} - 2\\ = 78 + 21 - 2\\ = 97\end{array}\)

Compute the critical value

Referring to the t-distribution table, use the column for 0.05 (Area in One Tail) and the row with 97 degrees of freedom.

Thus, the critical value is equal to 1.6607.

The p-value of t equal to 1.7048 is equal to 0.0456.

Conclusion of the test

For a right-tailed test, if the calculated value is greater than the critical value, reject the null hypothesis.

In this scenario, 1.7048 > 1.6607. Hence,reject the null hypothesis.

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.

Find the margin of error to compute the confidence interval

The confidence level will be equal to 90% if the hypothesis test is two-tailed at a 0.05 level of significance.

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

The margin of error is as follows.

\(\)\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \sqrt {\frac{{s_p^2}}{{{n_1}}} + \frac{{s_p^2}}{{{n_2}}}} \\ = 1.6607 \times \sqrt {\frac{{203.56}}{{78}} + \frac{{203.56}}{{21}}} \\ = 5.82509\end{array}\)

Compute the confidence interval

The value of the confidence interval is as follows.

\(\begin{array}{c}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < {\mu _1} - {\mu _2} < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\\left( {92.88462 - 86.90476} \right) - 5.82509 < {\mu _1} - {\mu _2} < \left( {92.88462 - 86.90476} \right) + 5.82509\\0.15 < {\mu _1} - {\mu _2} < 11.80\end{array}\)

90% of the confidence interval is equal to (0.15, 11.80).

Conclusion of the confidence interval

The interval does not contain the value of 0 and consists of only positive values. 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.

Comparison of the significance

The p-value of the test statistic obtained in exercise 12 is equal to 0.0168.

The p-value of the test statistic obtained above is equal to 0.0456.

The 90% confidence interval obtained in exercise 12 is equal to (1.46, 10.50).

The 90% confidence interval obtained above is equal to (0.15, 11.80).

As the p-value has increased and the confidence interval has become wider, it can be said that the assumption of equal population standard deviations lowers the significance of the test.