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Chapter 6: Problem 102

IQ tests scale the scores so that the mean IQ score is \(\mu=100\) and standard deviation is \(\sigma=15\). Suppose that 30 fourth graders in one class are given such an IQ test that is appropriate for their grade level. If the students are really a random sample of all fourth graders, what is the chance that the average IQ score for the class is above \(105 ?\)

Short Answer

Expert verified
The exact probability will depend on the calculated z-score and the precise values in the z-table, but the necessary steps involve calculating the z-score and then subtracting the probability associated with that z-score from 1.

Step by step solution

01

Calculate the z-score

A z-score represents how many standard deviations away from the mean a given point is. To find the z-score corresponding to an IQ score of \(105\), subtract the population mean from the target score and divide by the standard deviation. However, to find the standard deviation for class average, which is the standard deviation of the population divided by the square root of the sample size, this is \(\sigma = \frac{15}{\sqrt{30}}\). Thus, the z-score (\(z\)) can be calculated with the formula: \(z = \frac{105 - 100}{\sigma}\).
02

Calculate the probability

Once we have the z-score, we use a z-table (standard normal distribution table) to find the probability that the z-score is above our calculated value. A z-table gives the probability of a randomly chosen z-score being below a given value. So to get the probability of our z-score being greater than the calculated z-score, we'll subtract the probability found in the z-table from 1.

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Key Concepts

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

Z-Score
The concept of a z-score is fundamental in statistics, particularly when dealing with standardized test scores like IQ tests. A z-score tells us how far away a particular value is from the mean, measured in units of standard deviation. It's like a universal language for comparing different sets of data. To calculate a z-score, one subtracts the mean from the value in question and then divides that result by the standard deviation. In the context of our exercise where the mean IQ score is 100, and the standard deviation is 15, finding the z-score lets us understand how 'unusual' an average IQ score of 105 is for the class.
Standard Deviation
Understanding the standard deviation is crucial because it tells us how spread out the values in a set of data are. In the case of IQ scores, a standard deviation of 15 means that most people's scores are within 15 points of the mean, either lower or higher. However, when we're looking at the average score of multiple people, we actually use a different value called the standard error, which accounts for sample size. The smaller the standard error, the closer we expect the sample average to be to the population mean. This impacts how the z-score is calculated because it alters the denominator of the equation, and thus the resulting value and its interpretation.
Probability
Probability is essentially a way of expressing the likelihood of an event occurring. In statistics, we're often concerned with the probability of observing a certain value within a normal distribution. This is represented by a number between 0 and 1, or sometimes as a percentage. In the classroom IQ example, we're interested in the probability of the class average being above 105. This requires us to use our calculated z-score to find the corresponding likelihood from a z-table, which gives us the cumulative probability for values below a given z-score. By subtracting this from 1, we ascertain the chance of observing a class average above 105.
Normal Distribution
The normal distribution, also known as the bell curve, is a symmetrical distribution where most of the data points cluster around the mean, with values tapering off as they move away in either direction. It is the basis for many statistical procedures because many psychological tests, including IQ tests, assume that scores are normally distributed. This allows statisticians to make inferences about populations from sample data. When you hear about z-scores and probabilities, it's usually in reference to a normal distribution, and our exercise relies on this principle. This distribution is also crucial in calculating confidence intervals and for hypothesis testing.

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Most popular questions from this chapter

Rrefer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cardiovascular disease? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

A sample with \(n=75, \bar{x}=18.92,\) and \(s=10.1\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman having a fracture? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$
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