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Confidence Intervals: Sample Size A random sample is drawn from a population with \(\sigma=12\). The sample mean is 30 . (a) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 49 . What is the value of the margin of error? (b) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 100 . What is the value of the margin of error? (c) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 225 . What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a \(90 \%\) confidence interval decrease?

Short Answer

Expert verified
The margin of error decreases with increased sample size, leading to shorter confidence intervals.

Step by step solution

01

Understanding the Problem

We need to compute confidence intervals for the population mean, given a population standard deviation \( \sigma = 12 \), a sample mean \( = 30 \), and different sample sizes. Additionally, we'll calculate the margin of error for each confidence interval.
02

Using the Confidence Interval Formula

For a known population standard deviation, the confidence interval for the population mean is given by \( \bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \sigma \) is the population standard deviation, \( n \) is the sample size, and \( z \) is the z-score corresponding to the desired confidence level.
03

Determining the Z-Score

For a \(95\%\) confidence interval, the z-score is approximately \(1.96\). This value is standard for most \(95\%\) confidence intervals.
04

Calculate Confidence Interval and Margin of Error for n=49

\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} = 1.96 \cdot \frac{12}{\sqrt{49}} = 1.96 \cdot \frac{12}{7} = 1.96 \times 1.7143 \approx 3.36 \] Thus, the confidence interval is \( 30 \pm 3.36 \), or \([26.64, 33.36]\).
05

Calculate Confidence Interval and Margin of Error for n=100

\[ \text{Margin of Error} = 1.96 \cdot \frac{12}{\sqrt{100}} = 1.96 \cdot \frac{12}{10} = 1.96 \times 1.2 = 2.352 \] The confidence interval is \( 30 \pm 2.352 \), or \([27.648, 32.352]\).
06

Calculate Confidence Interval and Margin of Error for n=225

\[ \text{Margin of Error} = 1.96 \cdot \frac{12}{\sqrt{225}} = 1.96 \cdot \frac{12}{15} = 1.96 \times 0.8 = 1.568 \] The confidence interval is \( 30 \pm 1.568 \), or \([28.432, 31.568]\).
07

Comparing Margins of Error

The margins of error are: \(3.36\) for \(n=49\), \(2.352\) for \(n=100\), and \(1.568\) for \(n=225\). As the sample size increases, the margin of error decreases, supporting the principle that larger samples provide more precise estimates.
08

Critical Thinking on Confidence Interval Lengths

The lengths of the confidence intervals are: \(6.72\) for \(n=49\), \(4.704\) for \(n=100\), and \(3.136\) for \(n=225\). As the sample size increases, the confidence interval length decreases, further confirming that larger samples lead to more accurate and precise estimates.

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Key Concepts

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

Margin of Error
The margin of error is a crucial concept in statistics, particularly when dealing with confidence intervals. It represents the range around the sample mean within which the true population mean is likely to lie. In simpler terms, it tells us how much the sample mean might differ from the actual population mean due to random sampling error.
Understanding the margin of error allows you to grasp the uncertainty inherent in statistical estimates. It is calculated using the formula: \[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} \] where:
  • \( z \) is the z-score corresponding to the desired confidence level (e.g., for a 95% confidence interval, \( z = 1.96 \)).
  • \( \sigma \) is the population standard deviation, which measures the amount of variation in the data.
  • \( n \) is the sample size.
As the formula shows, the margin of error is directly influenced by the sample size and population standard deviation. A smaller margin of error indicates a more precise estimate of the population mean.
Sample Size
Sample size, or the number of observations in a sample, is a key component affecting the margin of error and the confidence interval. As seen in the provided exercise, changing the sample size significantly impacts the conclusions we can draw about the population.
A larger sample size generally results in a smaller margin of error. This is because with more data points, the estimates of the population parameters become more stable and reliable. Let's consider:
  • When the sample size increased from 49 to 225, the margin of error decreased.
  • Larger sample sizes reduce sampling variability, leading to narrower confidence intervals.
  • These narrower intervals provide more precision in estimating the population parameter.
By selecting an appropriate sample size, researchers can ensure that their statistical analyses yield meaningful and reliable results. Remember, while larger samples are desirable for accuracy, practical constraints such as cost and time may limit the feasible sample size.
Critical Thinking on Data
Critical thinking in the context of data and statistics involves analyzing and evaluating the information presented to make informed decisions. When looking at confidence intervals, critical thinking allows us to interpret what these intervals imply about the data and our conclusions.
For instance, as sample sizes increase, the length of the confidence interval decreases. This means our estimate of the population parameter becomes more precise. Here are key points to consider:
  • **Reasoning:** Understand why confidence intervals shrink with larger samples; it's due to reduced variability.
  • **Validity:** Evaluate the assumptions underlying the intervals; are they relevant and applicable to your data?
  • **Implications:** Recognize what narrow intervals suggest about the certainty of your estimates, adding confidence to your findings.
By enhancing your critical thinking skills, you can better assess the quality and implications of statistical analyses. This includes evaluating if broader decisions should be based on the data or if further investigation is needed.

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Most popular questions from this chapter

Expand Your Knowledge: Sample Size, Difference of Proportions What about the sample size \(n\) for confidence intervals for the difference of proportions \(p_{1}-p_{2}\) ? Let us make the following assumptions: equal sample sizes \(n=n_{1}=n_{2}\) and all four quantities \(n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2}\), and \(n_{2} \hat{q}_{2}\) are greater than \(5 .\) Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and a given maximal margin of error \(E\) for a specified confidence level \(c\), then the sample size \(n\) should be at least $$ n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right) $$ However, if we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\), then the theory similar to that used in this section tells us that the sample size \(n\) should be at least $$ n=\frac{1}{2}\left(\frac{z_{c}}{E}\right)^{2} $$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be \(99 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.04\). Use the preliminary estimates \(\hat{p}_{1}=289 / 375\) for the proportion of couples sharing two personality traits and \(\hat{p}_{2}=23 / 571\) for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )? (b) Suppose that in Problem 17 we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and we want to be \(95 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.05 .\) How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )?

Brain Teaser A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p>5\) and \(n q>5 .\) Since we usually do not know \(p\), we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p} .\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5\). Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed \(5 .\) Hint \(:\) In the inequality \(n \hat{p}>5\), replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5\), replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Brain Teaser: Algebra Why do we use \(1 / 4\) in place of \(p(1-p)\) in formula (22) for sample size when the probability of success \(p\) is unknown? (a) Show that \(p(1-p)=1 / 4-(p-1 / 2)^{2}\). (b) Why is \(p(1-p)\) never greater than \(1 / 4\) ?

Answer true or false. Explain your answer. If the sample mean \(\bar{x}\) of a random sample from an \(x\) distribution is relatively small, then the confidence interval for \(\mu\) will be relatively short.

Ecology: Sand Dunes At wind speeds above 1000 centimeters per second \((\mathrm{cm} / \mathrm{sec})\), significant sand-moving events begin to occur. Wind speeds below \(1000 \mathrm{~cm} / \mathrm{sec}\) deposit sand, and wind speeds above \(1000 \mathrm{~cm} / \mathrm{sec}\) move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes (Reference: Hydraulic, Geologic, and Biologic Research at Great Sand Dunes National Monument and Vicinity, Colorado, Proceedings of the National Park Service Research Symposium). At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty wind speed readings gave an average velocity of \(\bar{x}=1075 \mathrm{~cm} / \mathrm{sec} .\) Based on long-term experience, \(\sigma\) can be assumed to be \(265 \mathrm{~cm} / \mathrm{sec} .\) (a) Find a \(95 \%\) confidence interval for the population mean wind speed at this site. (b) Interpretation Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.

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