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Chapter 19: Problem 36

Infants weighing less than 1500 grams at birth are classed as "very low birth weight." Low birth weight carries many risks. One study followed 113 male infants with very low birth weight to adulthood. At age 20, the mean IQ score for these men was \(x^{-} \bar{x}=\) 87.6. \({ }^{5}\) IQ scores vary Normally with standard deviation \(\sigma=15\). Give a \(95 \%\) confidence interval for the mean IQ score at age 20 for all very-low-birth-weight males.

Short Answer

Expert verified
The 95% confidence interval is approximately [84.8, 90.4].

Step by step solution

01

Understand the Problem

We are given a sample mean IQ score and the population standard deviation, and we need to calculate a 95% confidence interval for the mean IQ score of very-low-birth-weight males at age 20. The sample size is 113.
02

Determine the Confidence Interval Formula

For the confidence interval of the mean when the standard deviation is known, use the formula: \( \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \). Here, \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
03

Find the z-score for 95% Confidence

For a 95% confidence interval, the z-score is approximately 1.96, because the standard normal distribution has 95% of the data within 1.96 standard deviations from the mean.
04

Calculate the Standard Error

Calculate the standard error of the mean using the formula: \(\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 15\) and \(n = 113\). So, Standard Error \( = \frac{15}{\sqrt{113}} \approx 1.411\).
05

Calculate Margin of Error

The margin of error is calculated by multiplying the z-score by the standard error: \(1.96 \times 1.411 \approx 2.765\).
06

Construct the Confidence Interval

Add and subtract the margin of error from the sample mean to find the confidence interval: \(87.6 \pm 2.765\). This yields a confidence interval of \([84.835, 90.365]\).

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Key Concepts

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

Standard Error
The concept of Standard Error is crucial when we want to estimate how much our sample mean varies from the actual population mean. The standard error gives us an idea of the accuracy of the sample mean as a representative of the population mean. It is calculated by dividing the population standard deviation by the square root of the sample size. This can be given by the formula: \[SE = \frac{\sigma}{\sqrt{n}}\]Where,
  • \( \sigma \) is the population standard deviation
  • \( n \) is the sample size
In our case, we calculated the standard error as approximately 1.411. The smaller the standard error, the closer the sample mean is to the population mean. This helps in determining how reliable our sample is in estimating the population.
Z-score
A Z-score is a measure that describes a value's relation to the mean of a group of values, expressed in terms of standard deviations. In the context of a confidence interval, the Z-score helps us determine the margin of error, which leads to the range that estimates the population parameter. For a 95% confidence level, the Z-score commonly used is 1.96. This is because 95% of the data under a normal distribution curve falls within 1.96 standard deviations from the mean. By finding the Z-score, we account for the variation and provide a range that is likely to include the actual population mean. This statistical tool ensures that we take into consideration the variation that may not be evident just from the sample.
Sample Mean
The sample mean is the average of observations in our sample, a crucial statistic when estimating the population mean. In the exercise we looked at, the sample mean IQ score was given as 87.6. This figure represents our best estimate of the true mean IQ score of all very-low-birth-weight males at age 20. By using this sample mean, along with the Z-score and standard error, we construct a confidence interval to estimate the range this true mean lies within. Calculating the sample mean involves simply averaging the IQ scores in our sample. The more observations we have, the more reliable our sample mean becomes as an estimate of the population mean. It's a simple yet powerful notion in statistics.
Population Standard Deviation
Population Standard Deviation (\( \sigma \)) measures the amount of variation or dispersion in a set of values in the entire population. It provides insight into how much individual data points deviate from the mean of the population.In our problem, the population standard deviation is given as 15, which stays constant regardless of the particular sample we analyze. This measure is crucial as it helps in calculating the standard error and consequently in forming the confidence interval. A higher standard deviation indicates more variability in the population, which can lead to wider confidence intervals. For estimating the mean IQ score, knowing the population standard deviation helps in understanding the overall spread of IQ scores among all very-low-birth-weight males.

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Most popular questions from this chapter

Americans spend more than \(\$ 30\) billion annually on a variety of weight loss products and services. In a study of retention rates of those using the Rewards Program at Jenny Craig in 2005, it was found that about \(18 \%\) of those who began the program dropped out in the first four weeks. \({ }^{14}\) Assume we have a random sample of 300 people beginning the program. (a) What is the mean number of people who would drop out of the Rewards Program within four weeks in a sample of this size? What is the standard deviation? (b) What is the probability that at least 235 people in the sample will still be in the Rewards Program after the first four weeks? Check that the Normal approximation is permissible and use it to find this probability. If your software allows, find the exact binomial probability and compare the two results.

Choose a college freshman at random and ask what type of high school he or she attended. Here is the distribution of results: 13 $$ \begin{array}{l|llllll} \hline \text { Type } & \begin{array}{l} \text { Regular } \\ \text { Public } \end{array} & \begin{array}{l} \text { Public } \\ \text { Charter } \end{array} & \begin{array}{l} \text { Public } \\ \text { Magnet } \end{array} & \begin{array}{l} \text { Private } \\ \text { Religious } \end{array} & \begin{array}{l} \text { Private } \\ \text { Independent } \end{array} & \begin{array}{l} \text { Home } \\ \text { School } \end{array} \\ \hline \text { Probability } & 0.758 & 0.029 & 0.035 & 0.109 & 0.063 & 0.006 \\\ \hline \end{array} $$ What is the conditional probability that a college freshman was home schooled, given that he or she did not attend a regular public high school?

Alysha makes \(40 \%\) of her free throws. She takes five free throws in a game. If the shots are independent of each other, the probability that she misses the first two shots but makes the other three is about (a) \(0.230\). (b) \(0.115\). (c) \(0.023\). (d) \(0.600\).

According to the National Survey of Student Engagement (NSSE), the average amount of time that first-year college students spent preparing for class (studying, reading, writing, doing homework or lab work, analyzing data, rehearsing, and other academic activities) in 2013 was \(14.25\) hours per week. Your college wonders if the average \(\mu\) for its first-year students in 2013 differed from the national average. A random sample of 500 students who were first-year students in 2013 claims to have spent an average of \(x^{-} \bar{x}=13.4\) hours per week on homework in their first year. What are the null and alternative hypotheses for a comparison of first-year students at your college with national first-year students in 2013 ? (a) \(H_{0}: \mathrm{x}^{-} \bar{x}=14.25, H_{a}: \mathrm{x}^{-} \bar{x} \neq 14.25\) (b) \(H_{0}: x^{-} \bar{x}=13.4, H_{a}: x^{-} \bar{x}>13.4\) (c) \(H_{0}: \mu=14.25, H_{a}: \mu \neq 14.25\) (d) \(H_{0}: \mu=13.4, H_{a}: \mu>13.4\)

Two wine tasters rate each wine they taste on a scale of 1 to 5. From data on their ratings of a large number of wines, we obtain the following probabilities for both tasters' ratings of a randomly chosen wine: $$ \begin{array}{cccccc} \hline &&& {\text { Taster 2 }} \\ \text { Taster 1 } & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 0.05 & 0.02 & 0.01 & 0.00 & 0.00 \\ \hline 2 & 0.02 & 0.08 & 0.04 & 0.02 & 0.01 \\ \hline 3 & 0.01 & 0.04 & 0.25 & 0.05 & 0.01 \\ \hline 4 & 0.00 & 0.02 & 0.05 & 0.18 & 0.02 \\ \hline 5 & 0.00 & 0.01 & 0.01 & 0.02 & 0.08 \\ \hline \end{array} $$ (a) Why is this a legitimate discrete probability model? (b) What is the probability that the tasters agree when rating a wine? (c) What is the probability that Taster 1 rates a wine higher than Taster 2? What is the probability that Taster 2 rates a wine higher than Taster 1 ?
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