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A Danish study of individuals born at a Copenhagen hospital between 1959 and 1961 reported higher mean IQ scores for adults who were breast-fed for longer lengths of time as babies (E. Mortensen et al., \(J A M A,\) vol. \(287,2002,\) pp. \(2365-2371\) ). The mean IQ score was \(98.1(s=15.9)\) for the \(272 \mathrm{sub}\) jects who had been breast-fed for no longer than a month and \(108.2(s=13.1)\) for the 104 subjects who had been breast-fed for seven to nine months. a. With software that can analyze summarized data, use an inferential method to analyze whether the corresponding population means differ, assuming the population standard deviations are equal. Interpret. b. Was this an experimental or an observational study? Can you think of a potential lurking variable?

Short Answer

Expert verified
a. The test shows a significant difference in mean IQ scores between the groups. b. It's an observational study; parental education could be a lurking variable.

Step by step solution

01

Identify Hypotheses

We are comparing the mean IQ scores for two groups: subjects breast-fed for no longer than a month and those breast-fed for seven to nine months. Our null hypothesis (H0) states that the population means of the two groups are equal, i.e., \( \mu_1 = \mu_2 \), where \( \mu_1 \) and \( \mu_2 \) are the means for the two groups, respectively. The alternative hypothesis (H1) claims that the means are different, i.e., \( \mu_1 eq \mu_2 \).
02

Choose Test and Calculate Test Statistic

Since we are assuming equal population standard deviations, we'll use a two-sample t-test for equal variances. The test statistic is calculated using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]where \( \bar{x}_1 = 98.1, \bar{x}_2 = 108.2, n_1 = 272, n_2 = 104 \). The pooled standard deviation \( s_p \) is found using: \[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\] Substitute the values: \( s_1 = 15.9 \) and \( s_2 = 13.1 \).
03

Calculate Pooled Standard Deviation

Calculate the pooled standard deviation: \[ s_p = \sqrt{\frac{(272 - 1) \cdot 15.9^2 + (104 - 1) \cdot 13.1^2}{272 + 104 - 2}} = \sqrt{\frac{(271 \cdot 252.81) + (103 \cdot 171.61)}{374}} \]which simplifies to \[ s_p = \sqrt{\frac{68411.51 + 17663.83}{374}} = \sqrt{229.89} = 15.16 \]
04

Compute t-Statistic

Substitute the pooled standard deviation back into the t-statistic formula:\[ t = \frac{98.1 - 108.2}{15.16 \cdot \sqrt{\frac{1}{272} + \frac{1}{104}}}\]which simplifies to:\[ t = \frac{-10.1}{15.16 \cdot 0.127} = \frac{-10.1}{1.925} = -5.25 \]
05

Determine p-value and Conclusion

Using the t-distribution table or software, find the p-value corresponding to \( t = -5.25 \) for degrees of freedom \( n_1 + n_2 - 2 = 374 \). The p-value is very small, indicating strong evidence against the null hypothesis. Thus, we reject \( H_0 \) and conclude that there is a significant difference in IQ scores between the two groups.
06

Classify Study Type and Identify Lurking Variable

This study is observational since it observes and measures outcomes without manipulating any variables. A potential lurking variable could be socioeconomic status or education level of the parents, which might influence both baby feeding choices and later IQ scores.

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Key Concepts

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

Two-Sample T-Test
A two-sample t-test is an inferential statistical method used when you want to compare the means of two independent groups. In this case, we compared the mean IQ scores of individuals who were breast-fed for varying lengths of time as babies. By using a two-sample t-test, researchers can determine if the difference between the means of these two groups is statistically significant, or if it might have happened just by chance.

The test begins by establishing two hypotheses: the null hypothesis, which assumes there is no difference in the group means, and the alternative hypothesis, which suggests the group means are not equal. For the Danish study, the null hypothesis \( H_0 \) is that the mean IQ for both groups is the same: \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_1 \) assumes they differ: \( \mu_1 eq \mu_2 \).

The calculation of the t-statistic involves the difference between the sample means divided by the pooled estimate of the standard deviation, which accounts for the size of both groups. A significantly large or small t-value supports the alternative hypothesis, suggesting a real difference in population means.
Observational Study
An observational study is a type of research method where the researcher observes subjects without manipulating any variables or conditions. In the Danish study, the researchers observed two groups of individuals based on their duration of breastfeeding and measured their IQ scores in adulthood.

This method contrasts with experimental studies where interventions are applied to one group, and outcomes are compared to a control group. Observational studies like this one rely on naturally occurring differences among subjects. They're valuable because they allow researchers to collect data over time without interference, but they can also be limited. Without controlled conditions, it's harder to establish causal relationships.

A potential "lurking variable" is a variable that might influence both the breastfeeding duration and the IQ outcomes but is not directly studied. Possible lurking variables in this study could include socioeconomic status or parental education levels. These factors could influence both the feeding practices and the educational opportunities, potentially affecting the child's IQ.
Pooled Standard Deviation
The pooled standard deviation \( s_p \) is a valuable statistic used to estimate the overall variability across two or more groups that we assume have equal variances. It provides a weighted average of the individual groups' standard deviations, reflecting the combined variability of the two groups.

To calculate the pooled standard deviation \( s_p \), you use the formula: \[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \] where \( n_1 \) and \( n_2 \) are the sample sizes, and \( s_1 \) and \( s_2 \) are the standard deviations of the two groups.

In the Danish study's context, the pooled standard deviation was calculated using the given sample sizes and standard deviations for the two breastfeeding groups. By having a combined measure of variability, it allows the t-test to be more reflective of the entire sample pool's standard deviation, providing a basis for further statistical analysis like the two-sample t-test. This approach assumes that both groups come from populations with the same variance, which is crucial for the validity of the t-test results.

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Most popular questions from this chapter

A test consists of 100 true-false questions. Joe did not study, and on each question he randomly guesses the correct response. Jane studied a little and has a 0.60 chance of a correct response for each question. a. Approximate the probability that Jane's score is nonetheless lower that Joe's. (Hint: Use the sampling distribution of the difference of sample proportions.) b. Intuitively, do you think that the probability answer to part a would decrease or increase if the test had only 50 questions? Explain.

A study of a sample of horseshoe crabs on a Florida island (J. Brockmann, Ethology, vol. \(102,1996,\) pp. \(1-21\) ) investigated the factors that were associated with whether female crabs had a male crab mate. Basic statistics, including the five-number summary on weight (kg) for the 111 female crabs who had a male crab nearby and for the 62 female crabs who did not have a male crab nearby, are given in the table. Assume that these horseshoe crabs have the properties of a random sample of all such crabs. $$\begin{array}{lcccccccc}\hline {\text { Summary Statistics for Weights of Horseshoe Crabs }} \\ \hline & n & \text { Mean } & \text { Std. Dev. } & \text { Min } & \text { Q1 } & \text { Med } & \text { Q3 } & \text { Max } \\\\\hline \text { Mate } & 111 & 2.6 & 0.6 & 1.5 & 2.2 & 2.6 & 3.0 & 5.2 \\ \text { No Mate } & 62 & 2.1 & 0.4 & 1.2 & 1.8 & 2.1 & 2.4 & 3.2 \\ \hline\end{array}$$ a. Sketch box plots for the weight distributions of the two groups. Interpret by comparing the groups with respect to shape, center, and variability. b. Estimate the difference between the mean weights of female crabs who have mates and female crabs who do not have mates. c. Find the standard error for the estimate in part b. d. Construct a \(90 \%\) confidence interval for the difference between the population mean weights, and interpret.

Refer to Example 10 on whether arthroscopic surgery is better than placebo. The following table shows the pain scores one year after surgery. Using software (such as MINITAB) that can conduct analyses using summary statistics, compare the placebo to the debridement group, using a \(95 \%\) confidence interval. Use the method that assumes equal population standard deviations. Explain how to interpret the interval found by using software. $$\begin{array}{lccc} \hline & & {\text { Knee Pain Score }} \\ { 3 - 4 } \text { Group } & \text { Sample Size } & \text { Mean } & \text { Standard Deviation } \\ \hline \text { Placebo } & 60 & 48.9 & 21.9 \\ \text { Arthroscopic }- & 61 & 54.8 & 19.8 \\\\\text { lavage } & & & \\ \begin{array}{l}\text { Arthroscopic }- \\ \text { debridement }\end{array} & 59 & 51.7 & 22.4 \\ \hline\end{array}$$

In the United States, the median age of residents is lowest in Utah. At each age level, the death rate from heart disease is higher in Utah than in Colorado. Overall, the death rate from heart disease is lower in Utah than Colorado. Are there any contradictions here, or is this possible? Explain.

A study of the death penalty in Kentucky reported the results shown in the table. (Source: Data from \(\mathrm{T}\). Keil and \(\mathrm{G}\). Vito, Amer. \(J\). Criminal Justice, vol. \(20,1995,\) pp. \(17-36 .)\) a. Find and compare the percentage of white defendants with the percentage of black defendants who received the death penalty, when the victim was (i) white and (ii) black. b. In the analysis in part a, identify the response variable, explanatory variable, and control variable. c. Construct the summary \(2 \times 2\) table that ignores, rather than controls, victim's race. Compare the overall percentages of white defendants and black defendants who got the death penalty (ignoring, rather than controlling, victim's race). Compare to part a. d. Do these data satisfy Simpson's paradox? If not, explain why not. If so, explain what is responsible for Simpson's paradox occurring. e. Explain, without doing the calculations, how you could inferentially compare the proportions of white and black defendants who get the death penalty (i) ignoring victim's race and (ii) controlling for victim's race. $$\begin{array}{llrrr}\hline \begin{array}{l} \text { Victim's } \\\\\text { Race } \end{array} & \begin{array}{l} \text { Defendant's } \\\\\text { Race } \end{array} & \begin{array}{l}\text { Death Penalty } \\\\\text { Yes } \end{array} & \begin{array}{r} \text { No } \end{array} & \text { Total } \\ \hline \text { White } & \text { White } & 31 & 360 & 391 \\ & \text { Black } & 7 & 50 & 57 \\ \text { Black } & \text { White } & 0 & 18 & 18 \\ & \text { Black } & 2 & 106 & 108 \\\\\hline\end{array}$$

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