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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-birth-weight 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 \(0.10\).

Short Answer

Expert verified
b. between 0.01 and 0.05.

Step by step solution

01

State the Hypotheses

We are testing if there is a statistically significant difference in mean IQ scores between the two groups. The null hypothesis (\( H_0 \)) is that there is no difference: \( \mu_1 = \mu_2 \). The alternative hypothesis (\( H_a \)) is that there is a difference: \( \mu_1 eq \mu_2 \).
02

Collect Data

We know from the problem that the VLBW group has \( n_1 = 38 \) with a mean IQ of \( \bar{x}_1 = 86.2 \) and a standard deviation of \( s_1 = 13.4 \). The control group has \( n_2 = 54 \) with a mean IQ of \( \bar{x}_2 = 89.8 \) and a standard deviation of \( s_2 = 14.0 \).
03

Calculate the Standard Error

To calculate the standard error (SE) of the difference between the two means, use the formula: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{13.4^2}{38} + \frac{14.0^2}{54}}.\]
04

Calculate the Test Statistic

Use the test statistic for two independent means:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{86.2 - 89.8}{SE}.\]
05

Determine Degrees of Freedom

For two independent samples, approximate the degrees of freedom (df) using:\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{ \frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}}.\]
06

Calculate the P-value

Using the calculated \( t \)-value and degrees of freedom, look up or use software to determine the two-tailed P-value from the student's t-distribution.
07

Decision Based on P-value

Compare the P-value to the significance level thresholds. If the P-value is less than 0.01, it is statistically significant at the 1% level; if between 0.01 and 0.05, then significant at the 5% level. If between 0.05 and 0.10, marginally significant; greater than 0.10 is not significant.

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Key Concepts

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

Statistical Significance
Statistical significance tells us whether the results of our hypothesis test are not likely to be caused by chance alone. When comparing two groups, as done in this exercise with VLBW women and control group women, we check if the observed difference in their IQ scores is unlikely under the assumption that there's no real difference (the null hypothesis).

Here's how it works: When you perform a test, you calculate a p-value which measures the strength of the evidence against the null hypothesis. If this p-value is less than a set threshold (commonly 0.05, known as the significance level), you can claim with confidence that there's a statistically significant difference. This means the difference is not due to random chance, but likely reflects a true effect.

In simpler terms, determining statistical significance is like checking if your experimental finding is just a fluke or if it points to a genuine difference.
  • A p-value < 0.01 means very strong evidence against the null hypothesis.
  • Between 0.01 and 0.05 suggests strong evidence against the null.
  • 0.05 to 0.10 indicates weak or marginal evidence.
  • Greater than 0.10 means the results are not significant, or likely due to random variation.
Student's t-distribution
The Student's t-distribution is often used in statistics, especially when dealing with small sample sizes. It acts as a vital tool to determine how different the sample mean is from the population mean when the population standard deviation is unknown.

It's a family of distributions that look quite similar to the normal distribution but have heavier tails. What this means is that it provides a more conservative estimate of the mean when sample sizes are small, accommodating more variability which could be expected due to sampling error.

For this test involving IQ scores of the VLBW women and the control group women, the t-distribution helps us conduct a 't-test'. It evaluates the precision of the results and helps in making conclusions about the population from where samples were drawn, despite the limited sample set involved. As the sample size grows, the t-distribution becomes more like the normal distribution.

Understanding when to apply the Student's t-distribution is crucial:
  • Use it when the population standard deviation is unknown.
  • It's suitable for small sample sizes (n < 30 is a common guideline).
  • Aids hypothesis testing in assessing mean differences.
Standard Error
The standard error (SE) is crucial in hypothesis testing as it measures the variability of the sample mean. It gives us an idea of how much the sample mean is expected to vary from the true population mean.

In this particular exercise, we're interested in the standard error of the difference between two means, one from the VLBW group and another from the control group. The standard error helps quantify the uncertainty or how "spread out" our sample means are.

The formula for the standard error difference involves the standard deviations and sizes of both groups: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Here, \(s_1\) and \(s_2\) are standard deviations, and \(n_1\) and \(n_2\) are sample sizes for the two groups.

The smaller the standard error, the more reliable our sample mean is as an estimate of the population mean. It's a key component in calculating the test statistic, which ultimately helps determine the p-value.
  • SE reflects how much your sample mean is expected to fluctuate from the actual mean.
  • Lower SE suggests higher precision and reliability in your sample estimate.
  • It's vital in forming confidence intervals and hypothesis tests.
Degrees of Freedom
Degrees of freedom (df) are essential to the statistical analysis when calculating test statistics like the t-value. They provide a way to account for the sample size and the number of estimates or parameters being used in the hypothesis test.

The idea of degrees of freedom relates to the number of "free" values in a dataset that are free to vary, given certain constraints. In a two-sample test like this one, degrees of freedom are used to adjust the t-distribution, making it flexible enough to account for different sample sizes and variances.

For our exercise, the degrees of freedom for two independent samples can be calculated using the formula:\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{ \frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}}\]This approach gives a more accurate approximation by considering the individual variances and sizes of the samples.

Understanding degrees of freedom helps in determining the right t-distribution to use, which changes as the sample size and variance estimates change.
  • Degrees of freedom impact the shape of the t-distribution.
  • It adjusts the test statistic to consider sample size and variability.
  • More degrees of freedom usually indicate a more precise estimate.

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Most popular questions from this chapter

Case for the Supreme Court. In 1986, a Texas jury found a Black man guilty of murder. The prosecutors had used "peremptory challenges" to remove 10 of the 11 Blacks and four of the 31 Whites in the pool from which the jury was chosen. \(\frac{24}{}\) The law says that there must be a plausible reason (that is, a reason other than race) for different treatment of Blacks and Whites in the jury pool. When the case reached the Supreme Court 17 years later, the Court said that "happenstance is unlikely to produce this disparity." The inferential methods we have studied cannot safely be used to support the Court's finding that chance is unlikely to produce so large a Black-White difference. Why not?

Take \(p_{V L B W}\) and \(p_{\text {contrnd }}\) to be the proportions of all VLBW and normal-birth-weight (control) babies who would graduate from high school. The hypotheses to be tested are a. \(H_{0}: p_{V L B W}=p_{\text {control }}\) versus \(H_{a}: p_{V L B W} \neq p_{\text {control. }}\) b. \(H_{0}: p_{V L B W}=p_{\text {control }}\) versus \(H_{a}: p_{V L B W}>p_{\text {control. }}\) c. \(H_{0}: p_{V L B W}=p_{\text {control }}\) versus \(H_{a}: p_{V L B W}p_{\text {control }}\) versus \(H_{a}: p_{V L B W}=p_{\text {controi }}\)

Dyeing Fabrics. Different fabrics respond differently when dyed. This matters to clothing manufacturers, who want the color of fabric to match their specifications closely. A researcher dyed fabrics made of cotton and of ramie with the same "procion blue" dye applied in the same way. Then she used a colorimeter to measure the lightness of the color on a scale in which black is 0 and white is 100 . Here are the data for eight pieces of each fabric: 20 FBCDYE \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline Cotton & \(49.82\) & \(49.88\) & \(49.98\) & \(49.04\) & \(48.68\) & \(49.34\) & \(48.75\) \\ \hline & & & \(49.12\) \\ \hline \end{tabular} \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|} Ramie & \(41.72\) & \(41.83\) & \(42.05\) & \(41.44\) & \(41.27\) & \(42.27\) & \(41.12\) & \(41.49\) \\ \hline \end{tabular} Is there a significant difference between the fabrics? Which fabric is darker when dyed in this way?

A \(90 \%\) confidence interval for the difference in mean bone mineral density after 12 months on an alkalinizing diet and an acidifying diet is (use conservative Option 2 for the degrees of freedom) a. \(0.01\) to \(0.05\). b. \(-0.03\) to \(0.05\) c. \(-0.05\) to \(0.07\). d. \(-0.07\) to \(0.09\).

A 90 \%\( confidence interval for the mean bone mineral density of cats after 12 months on an acidifying diet is a. \)0.622\( to \)0.638\(. b. \)0.618\( to \)0.642\(. c. \)0.614\( to \)0.646\(. d. \)0.620\( to \)0.640$.

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