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A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(18.4,\) and the sample standard deviation, \(s\), is found to be \(4.5 .\) (a) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 35 (b) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(50 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(99 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(35 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=15,\) what conditions must be satisfied to compute the confidence interval?

Short Answer

Expert verified
Part (a): (16.909, 19.891). Part (b): (17.154, 19.646). Part (c): (16.440, 20.360). Higher sample size decreases margin of error, higher confidence level increases it. Part (d): Data should be nearly normal.

Step by step solution

01

- Understanding the confidence interval formula

The formula for the confidence interval for the mean is given by \[\bar{x} \pm E\]. The margin of error \(E\) is calculated as \(E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\) for large samples (or \(t_{\alpha/2}\) for small samples).
02

- Identifying values for part (a)

For part (a), the given values are: \(\bar{x} = 18.4\), \(s = 4.5\), \(n = 35\), and the confidence level is 95%, meaning \(\alpha = 0.05\).
03

- Calculating critical value for part (a)

For a 95% confidence level, the critical value (\(z_{\alpha/2}\)) is approximately 1.96 (from the standard normal distribution table).
04

- Calculating the margin of error for part (a)

The margin of error (\(E\)) is calculated as: \[E = 1.96 \cdot \frac{4.5}{\sqrt{35}} \approx 1.491\].
05

- Constructing the confidence interval for part (a)

The 95% confidence interval is then: \[18.4 \pm 1.491\] which is \[ (16.909, 19.891) \].
06

- Identifying values for part (b)

For part (b), the given values change to: \(\bar{x} = 18.4\), \(s = 4.5\), \(n = 50\), and the confidence level remains at 95%.
07

- Calculating the margin of error for part (b)

The margin of error for part (b) is: \[E = 1.96 \cdot \frac{4.5}{\sqrt{50}} \approx 1.246\].
08

- Constructing the confidence interval for part (b)

The 95% confidence interval is then: \[18.4 \pm 1.246\] which is \[ (17.154, 19.646) \].
09

- Impact of increasing sample size

Increasing the sample size decreases the margin of error. For part (a), \(E = 1.491\), and for part (b), \(E = 1.246\). Thus, a larger sample size results in a more precise estimate.
10

- Identifying values for part (c)

For part (c), the given values are: \(\bar{x} = 18.4\), \(s = 4.5\), \(n = 35\), and the confidence level is 99%, meaning \(\alpha = 0.01\).
11

- Calculating critical value for part (c)

For a 99% confidence level, the critical value (\(z_{\alpha/2}\)) is approximately 2.576 (from the standard normal distribution table).
12

- Calculating the margin of error for part (c)

The margin of error for part (c) is: \[E = 2.576 \cdot \frac{4.5}{\sqrt{35}} \approx 1.960\].
13

- Constructing the confidence interval for part (c)

The 99% confidence interval is then: \[18.4 \pm 1.960\] which is \[ (16.440, 20.360) \].
14

- Impact of increasing confidence level

Increasing the confidence level increases the margin of error. For part (a), \(E = 1.491\), and for part (c), \(E = 1.960\). Thus, a higher confidence level results in a wider interval.
15

- Conditions for computing confidence interval in part (d)

For small sample sizes like \(n = 15\), the data should be nearly normally distributed, as the Central Limit Theorem may not apply. This can be checked using graphical techniques (histograms, Q-Q plots) or statistical tests of normality.

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Key Concepts

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

Confidence Level
The confidence level represents the probability that the true population parameter lies within the confidence interval. It is usually expressed as a percentage, such as 95% or 99%.

When constructing confidence intervals, choosing a higher confidence level (e.g., 99% instead of 95%) means you are more certain that the interval contains the population mean. However, a higher confidence level also results in a wider confidence interval since you're incorporating more of the possible sample variations.
Margin of Error
The margin of error (E) measures the extent of the range around the sample mean within which the true population mean is expected to lie. It provides an estimate of the uncertainty associated with the sample data.

Mathematically, the margin of error is calculated as \(E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\), where \(z_{\alpha/2}\) is the critical value, s is the sample standard deviation, and n is the sample size. Therefore, E depends on the critical value, the standard deviation, and the size of the sample.
Sample Size
Sample size (n) significantly impacts the width of the confidence interval. A larger sample size generally reduces the margin of error, making the confidence interval narrower and providing a more precise estimate of the population mean.

This relationship is mathematically shown in the equation for the margin of error: \(E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\). As the sample size increases, the denominator (\(\sqrt{n}\)) increases, resulting in a smaller E and thus a narrower confidence interval.
Critical Value
The critical value is a key component in calculating the margin of error and therefore the confidence interval. It indicates the number of standard deviations you must go out from the mean to encompass a certain percentage of the data. For example, for a 95% confidence level, the critical value (\(z_{\alpha/2}\)) is approximately 1.96, taken from the standard normal distribution table.

A higher confidence level requires a larger critical value, indicating that you must stretch further from the sample mean to include more of the sample data in the interval, hence resulting in a wider interval.
Central Limit Theorem
The Central Limit Theorem (CLT) is crucial when constructing confidence intervals, especially for larger sample sizes. It states that the distribution of the sample mean will approximate a normal distribution, regardless of the shape of the original population distribution, if the sample size is sufficiently large (typically n > 30).

This theorem allows us to use normal distribution properties, such as critical values from the z-distribution, to construct confidence intervals even if the underlying population distribution is not normal. For small sample sizes, the data should ideally be normally distributed to apply these methods accurately.

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Most popular questions from this chapter

Home Ownership An urban economist wishes to estimate the proportion of Americans who own their house. What size sample should be obtained if he wishes the estimate to be within 0.02 with 90\% confidence if (a) he uses an estimate of 0.675 from the fourth quarter of 2000 obtained from the U.S. Census Bureau? (b) he does not use any prior estimates?

A Gallup poll conducted May \(20-22,2005,\) asked 1006 Americans, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that \(\bar{x}=13.4\) books and \(s=16.6\) books. Construct a \(99 \%\) confidence interval for the mean number of books Americans read either all or part of during the preceding year. Interpret the interval.

A Gallup poll conducted January \(23,2003-\) February \(10,2003,\) asked American teens (aged 13 to 17 ) how much time they spent each week using the Internet. How many subjects are needed to estimate the time American teens spend on the Internet each week within 0.5 hour with \(95 \%\) confidence? Initial survey results indicate that \(\sigma=6.6\) hours.

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?

Construct the appropriate confidence interval. A simple random sample of size \(n=785\) adults was asked if they follow college football. Of the 785 surveyed, 275 responded that they did follow college football. Construct a \(95 \%\) confidence interval about the population proportion of adults who follow college football.

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