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Chapter 9: Problem 29

IQ scores based on the Wechsler Intelligence Scale for Children (WISC) are known to be normally distributed with \(\mu=100\) and \(\sigma=15\) (a) Simulate obtaining 20 samples of size \(n=15\) from this population. (b) Obtain the sample mean and standard deviation for each of the 20 samples. (c) Construct \(95 \%\) t-intervals for each of the 20 samples. (d) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

Short Answer

Expert verified
Expect 19 intervals to contain the population mean. Verify with actual calculations.

Step by step solution

01

- Simulating 20 Samples

Generate 20 samples from a normal distribution with mean \(\mu=100\) and standard deviation \(\sigma=15\). Each sample should have size \(n=15\). This can be done using statistical software or programming languages such as Python or R.
02

- Calculate Sample Means

For each of the 20 samples, calculate the sample mean \(\bar{x}\). The sample mean is given by \[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\].
03

- Calculate Sample Standard Deviations

For each of the 20 samples, calculate the sample standard deviation \(s\). The sample standard deviation is given by \[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}\].
04

- Construct 95% t-Intervals

For each sample, construct the 95% t-interval using the formula \[ \bar{x} \pm t_{\frac{\alpha}{2}, df} \frac{s}{\sqrt{n}} \], where \(t_{\frac{\alpha}{2}, df}\) is the critical value from the t-distribution with \(df = n - 1\).
05

- Calculate Expected Interval Coverage

Since each interval is a 95% confidence interval, we expect \(0.95 \times 20 = 19\) intervals to include the population mean \(\mu=100\).
06

- Count Actual Interval Coverage

Check how many of the constructed intervals actually contain the population mean \(\mu=100\). This involves checking the intervals whether \(100\) lies within the lower and upper bounds for each interval.

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Key Concepts

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

Normal Distribution
The term 'normal distribution' refers to a bell-shaped curve that is symmetrical about the mean. In the case of IQ scores, if you were to plot all IQ scores on a graph, most people鈥檚 scores would cluster around the average, which is usually 100. This means that the frequency of people having extremely high or low IQ scores drops as you move away from the mean. The normal distribution is defined by two parameters: \( \mu \) (the mean) and \(\sigma \) (the standard deviation). In our example, IQ scores have a \(\mu = 100 \) and \(\sigma = 15 \).
Sample Mean
The sample mean or \( \bar{x} \) is simply the average value of a set of sample data. When you take a number of observations from a population, the sample mean estimates the population mean \(\mu\). To calculate the sample mean, you add up all the values in your sample and then divide by the number of values. Mathematically, it's given by \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \. \] In our case, each sample consists of 15 IQ scores. By averaging these 15 scores, you get the sample mean, which serves as an estimate of the population mean.
T-Interval
When you want to estimate the mean of the population but you only have a small sample size, you use a t-interval. A t-interval is a range that likely contains the population mean. In the context of our problem, the 95% t-interval is used. This means we are 95% confident that the interval contains the true population mean \(\mu = 100\). The formula for the t-interval is: \[ \bar{x} \pm t_{\frac{\alpha}{2}, df} \frac{s}{\sqrt{n}} , \] where t-values come from a t-distribution and depend on the sample size, denoted as degrees of freedom (df). For each of our 20 samples, this formula helps us build our 95% confidence intervals.
Confidence Interval
A confidence interval gives a range that is likely to contain a population parameter, like the mean. In our IQ score example, each of our t-intervals provides an estimate of the range where the true population mean (100) lies. A 95% confidence interval implies that if we repeated our experiment many times, we expect 95% of those intervals to capture the true mean. This is why we expect approximately 19 out of our 20 samples' intervals to include the population mean. Whether an interval contains the mean is verified by checking if the mean (100) falls within the range of the upper and lower bounds of the interval.

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Most popular questions from this chapter

Death Penalty In a Harris Poll conducted in July 2000 , \(64 \%\) of the people polled answered yes to the following question: "Do you believe in capital punishment, that is, the death penalty, or are you opposed to it?" The margin of error in the poll was \(\pm 3 \%\) = 卤0.03 , and the estimate was made with \(95 \%\) confidence. How many people were surveyed?

A simple random sample of size \(n\) is drawn from a population that is normally distributed with population standard deviation, \(\sigma,\) known to be \(13 .\) The sample mean, \(\bar{x},\) is found to be 108. (a) Compute the \(96 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 25 (b) Compute the \(96 \%\) confidence interval about \(\mu\) if the sample size, \(n\), is \(10 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Compute the \(88 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(25 .\) Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the size of the margin of error, \(E ?\) (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why? (e) Suppose an analysis of the sample data revealed three outliers greater than the mean. How would this affect the confidence interval?

In a random sample of 100 estate tax returns that was audited by the Internal Revenue Service, it was determined that the mean amount of additional tax owed was \(\$ 3137 .\) Assuming the population standard deviation of the additional amount owed is \(\$ 2694,\) construct and interpret a \(90 \%\) confidence interval for the mean additional amount of tax owed for estate tax returns.

Construct the appropriate confidence interval. A simple random sample of size \(n=300\) individuals who are currently employed is asked if they work at home at least once per week. Of the 300 employed individuals surveyed, 35 responded that they did work at home at least once per week. Construct a \(99 \%\) confidence interval about the population proportion of employed individuals who work at home at least once per week.

Suppose the following small data set represents a simple random sample from a population whose mean is 50 and standard deviation is \(10 .\) $$\begin{array}{llllll}43 & 63 & 53 & 50 & 58 & 44 \\\\\hline 53 & 53 & 52 & 41 & 50 & 43\end{array}$$ (a) A normal probability plot indicates the data come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set, assuming \(\sigma=10\) (b) Suppose the observation, \(41,\) is inadvertently entered into the computer as \(14 .\) Verify that this observation is an outlier. (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50 and standard deviation is \(10 .\) $$\begin{array}{|llllll}43 & 63 & 53 & 50 & 58 & 44 \\\\\hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\\\\hline 49 & 56 & 57 & 50 & 38 & 42 \\\\\hline 59 & 54 & 57 & 41 & 63 & 37 \\\\\hline 46 & 54 & 42 & 48 & 53 & 41\end{array}$$ Verify that the sample mean for the large data set is the same as the sample mean for the small data set. (e) Compute a \(95 \%\) confidence interval for the large data set, assuming \(\sigma=10 .\) Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose the last observation, \(41,\) is inadvertently entered as \(14 .\) Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier, assuming \(\sigma=10 .\) Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?
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