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Chapter 24: Problem 28

IQ scores were available for 113 men in the VLBW group and for 106 men in the control group. The mean IQ for the 113 men in the VLBW group was 87.6, and the standard deviation was \(15.1\). The 106 men in the control group had mean IQ 94.7, with standard deviation 14.9. Is there good evidence that mean IQ is lower among VLBW men than among controls from similar backgrounds? To test this with a two-sample \(t\) test, the test statistic would be (a) \(t=-1.72\). (b) \(t=-3.50\). (c) \(t=-5.00\). (d) \(t=-7.10\).

Short Answer

Expert verified
The test statistic is approximately \( t = -3.50 \).

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis is that there is no difference in the mean IQ scores between VLBW men and control men, i.e., \( H_0: \mu_{VLBW} = \mu_{control} \). The alternative hypothesis is that the mean IQ score for VLBW men is lower, i.e., \( H_a: \mu_{VLBW} < \mu_{control} \).
02

Calculate the Test Statistic

To perform a two-sample t-test, use the formula:\[t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \( \bar{X}_1 = 87.6 \), \( \bar{X}_2 = 94.7 \), \( s_1 = 15.1 \), \( s_2 = 14.9 \), \( n_1 = 113 \), and \( n_2 = 106 \). Substitute these values into the formula to calculate \( t \).
03

Perform the Calculations

First, calculate the difference in means:\[\bar{X}_1 - \bar{X}_2 = 87.6 - 94.7 = -7.1\]Next, calculate the standard error:\[SE = \sqrt{\frac{15.1^2}{113} + \frac{14.9^2}{106}} = \sqrt{\frac{228.01}{113} + \frac{222.01}{106}}\]Calculate each term separately:\[\frac{228.01}{113} \approx 2.0186 \quad \text{and} \quad \frac{222.01}{106} \approx 2.0944\]Combine these results:\[SE = \sqrt{2.0186 + 2.0944} \approx \sqrt{4.113} \approx 2.0278\]Now, calculate the t-statistic:\[t = \frac{-7.1}{2.0278} \approx -3.50\]
04

Conclusion

The calculated t-statistic is approximately \(-3.50\). Thus, the correct answer is option (b) \( t = -3.50 \). This provides evidence that the mean IQ is lower in the VLBW group compared to the control group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Null Hypothesis
The null hypothesis (denoted as \(H_0\)) is a foundational concept in statistical testing. It represents the assumption that there is no significant difference or effect. In the context of our exercise, the null hypothesis states that there is no difference in the mean IQ scores between the Very Low Birth Weight (VLBW) men and the control group men. This can be mathematically expressed as \( \mu_{VLBW} = \mu_{control} \). To test this hypothesis, we need to gather evidence to determine if there is enough reason to reject it and accept the alternative. Typically, evidence is gathered using data from experiments or studies that, if compelling enough, allow us to conclude that any observed difference is not due to random chance.
Exploring the Alternative Hypothesis
The alternative hypothesis (denoted as \(H_a\)) is considered when we suspect that there may be a significant effect or difference. It serves as a counterpoint to the null hypothesis. In our specific context, the alternative hypothesis claims that the mean IQ score for VLBW men is lower than that of the control group. This is expressed as \( \mu_{VLBW} < \mu_{control} \). The purpose of a hypothesis test is to evaluate whether to reject the null hypothesis in favor of the alternative, based on the data. If the data show a statistically significant difference, then we might conclude that the alternative hypothesis can be considered possible.
What is a t-statistic?
The t-statistic is a crucial part of a two-sample t-test. It determines how many standard deviations our sample mean is from the null hypothesis mean. In simple terms, it measures the size of the difference relative to the variation in our sample data.We use the following formula to calculate the t-statistic for a two-sample test:
  • \( t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)
Where:
  • \( \bar{X}_1 \) and \( \bar{X}_2 \) are the sample means of the two groups.
  • \( s_1 \) and \( s_2 \) are the standard deviations of the groups.
  • \( n_1 \) and \( n_2 \) are the sample sizes.
To conclude the hypothesis test, we compare the calculated t-statistic to a critical value from the t-distribution. If it falls into a specific region, it implies significant results, supporting the alternative hypothesis.
Understanding Mean Difference
The mean difference is simply the difference between the average values of two groups. It acts as an estimator of the difference in population means. In our IQ score exercise, the mean difference between the VLBW group and the control group is calculated as follows:\[ \bar{X}_1 - \bar{X}_2 = 87.6 - 94.7 = -7.1 \]This mean difference of \(-7.1\) suggests that on average, men in the VLBW group have an IQ that is 7.1 points lower than those in the control group. When subjected to statistical testing in context with the variability and size of each sample, this mean difference helps determine if the observed difference is statistically significant.

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Most popular questions from this chapter

In 2006, the pharmaceutical company Merck released a vaccine named Gardasil for human papilloma virus, the most common cause of cervical cancer in young women. The Merck website gives results from "four placebo-controlled, double- blind, randomized clinical studies" with women 16 to 26 years of age, as follows: \({ }^{16}\) $$ \begin{array}{lrr|cc} \hline & n & \text { Cervical Cancer } & n & {\text { Genital Warts }} \\ \hline \text { Gardasil } & 8487 & 0 & 7897 & 1 \\ \text { Placebo } & 8460 & 32 & 7899 & 91 \\ \hline \end{array} $$ (a) Give a 99\% confidence interval for the difference in the proportions of young women who develop cervical cancer with and without the vaccine. (b) Do the same for the proportions who develop genital warts. (c) What do you conclude about the overall effectiveness of the vaccine?

A \(95 \%\) confidence interval for the mean score on the creativity test for those subjects who cheated is (a) \(3.60 \pm 0.19\). (b) \(3.60 \pm 0.39\). (c) \(3.60 \pm 2.02\). (d) \(1.27 \pm 0.43\).

Drowning in bathtubs is a major cause of death in children less than five years old. A random sample of parents was asked many questions related to bathtub safety. Overall, \(85 \%\) of the sample said they used baby bathtubs for infants. Estimate the percent of all parents of young children who use baby bathtubs.

A study of genetic influences on diabetes compared normal mice with similar mice genetically altered to remove a gene called \(a P 2\). Mice of both types were allowed to become obese by eating a high-fat diet. The researchers then measured the levels of insulin and glucose in their blood plasma. Here are some excerpts from their findings. \({ }^{24}\) The normal mice are called "wild-type" and the altered mice are called " \(a P 2^{-1-. "}\) Each value is the mean \(\pm\) SEM of measurements on at least 10 mice. Mean values of each plasma component are compared between \(a P 2^{-/-}\)mice and wild-type controls by Student's \(t\) test \((* P<0.05\) and \(* * P<0.005)\). $$ \begin{array}{lcc} \hline \text { Parameter } & \text { Wild Type } & \mathrm{aP2^{-/- }} \\ \hline \text { Insulin (ng/mL) } & 5.9 \pm 0.9 & 0.75 \pm 0.2^{* *} \\ \hline \text { Glucose (mg/dL) } & 230 \pm 25 & 150 \pm 17^{*} \\ \hline \end{array} $$ Despite much greater circulating amounts of insulin, the wild-type mice had higher blood glucose than the \(a P 2^{-1}\) - animals. These results indicate that the absence of aP2 interferes with the development of dietary obesityinduced insulin resistance. Other biologists are supposed to understand the statistics reported so tersely. (a) What does "SEM" mean? What is the expression for SEM based on \(n, \mathrm{x}^{-} \bar{x}\), and s from a sample? (b) Which of the tests we have studied did the researchers apply? (c) Explain to a biologist who knows no statistics what \(P<0.05\) and \(P<0.005\) mean. Which is stronger evidence of a difference between the two types of mice?

Of the 126 women in the VLBW group, 38 said they had used illegal drugs; 54 of the 124 control group women had done so. The IQ scores for the VLBW women who had used illegal drugs had mean \(86.2\) (standard deviation 13.4), and the normal-birthweight controls who had used illegal drugs had mean IQ \(89.8\) (standard deviation 14.0). Is there a statistically significant difference between the two groups in mean IQ? The \(P\)-value for this test is (a) less than \(0.01\). (b) between \(0.01\) and \(0.05\). (c) between \(0.05\) and \(0.10\). (d) greater than \(0.10\).
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