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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
(a) Margin of error is 3.36. (b) Margin of error is 2.35. (c) Margin of error is 1.57. Margin of error decreases as sample size increases. Confidence interval length decreases with larger samples.

Step by step solution

01

Understand the Confidence Interval Formula

To construct a confidence interval for the population mean \( \mu \), we use the formula: \[ \text{CI} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] where \( \bar{x} \) is the sample mean, \( z \) is the z-score corresponding to the desired confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. For a 95% confidence level, \( z \approx 1.96 \).
02

Compute (a) for Sample Size n=49

For sample size \( n = 49 \), the sample mean \( \bar{x} = 30 \), \( \sigma = 12 \), and \( z = 1.96 \). Compute the standard error \( SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{49}} = \frac{12}{7} \approx 1.714 \). The margin of error \( E = z \times SE = 1.96 \times 1.714 \approx 3.36 \). Thus, the 95% confidence interval is \( 30 \pm 3.36 \), which is \( (26.64, 33.36) \).
03

Compute (b) for Sample Size n=100

For sample size \( n = 100 \), compute \( SE = \frac{12}{\sqrt{100}} = \frac{12}{10} = 1.2 \). The margin of error \( E = 1.96 \times 1.2 \approx 2.35 \). Therefore, the 95% confidence interval is \( 30 \pm 2.35 \), which is \( (27.65, 32.35) \).
04

Compute (c) for Sample Size n=225

For sample size \( n = 225 \), compute \( SE = \frac{12}{\sqrt{225}} = \frac{12}{15} = 0.8 \). The margin of error \( E = 1.96 \times 0.8 \approx 1.57 \). Thus, the 95% confidence interval is \( 30 \pm 1.57 \), which is \( (28.43, 31.57) \).
05

Analysis of Margin of Error

The margins of error calculated are approximately 3.36, 2.35, and 1.57 for \( n = 49, 100, \) and \( 225 \) respectively. We observe that as the sample size increases, the margin of error decreases. This is because the standard error decreases as more data points are sampled, reducing the uncertainty around the population mean estimate.
06

Critical Thinking on Confidence Interval Length

The lengths of the confidence intervals decrease as the sample size increases: \( 6.72 \), \( 4.7 \), and \( 3.14 \) for \( n = 49, 100, \) and \( 225 \) respectively. Increasing the sample size reduces the width of the interval, indicating a more precise estimate of the population mean. This principle holds for any confidence level, including 90%, although the exact intervals would be calculated using a different z-score.

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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 in a confidence interval is crucial because it tells us how much the sample mean might differ from the true population mean. When a statistician or researcher is working with data, they want to provide an estimated range or interval within which they believe the actual population mean lies. This interval is built around the sample mean, with the margin of error adding a buffer in both directions.
The formula for calculating the margin of error is:
  • Margin of Error = z-score × Standard error
The higher the z-score or the standard error, the larger the margin of error will be. A large margin of error indicates more uncertainty in the estimated range, while a smaller margin of error suggests greater precision.
In the example provided, as the sample size increased from 49 to 225, the margin of error decreased, signifying increased confidence in the estimated range for the population mean.
Sample Size
The sample size is the number of observations or data points collected from the population to make statistical inferences. Increasing the sample size generally results in more reliable and precise estimates. This happens because a larger sample size reduces the variability associated with the sample mean.
In essence, the sample size impacts both the margin of error and the width of the confidence interval. As seen in the exercise, as the sample size increases, the margin of error decreases, indicating a smaller gap between the sample means and the population mean. In practical terms, researchers aim to collect as large a sample as feasible, balancing the resources available with the need for precision. However, doubling the sample size does not halve the margin of error, due to the square root relationship in the formula.
Standard Error
The standard error measures the dispersion or variability of the sample mean. It is derived from the population standard deviation and the sample size and is a key component in constructing a confidence interval.The formula for the standard error (SE) is:
  • SE = \( \frac{\sigma}{\sqrt{n}} \)
where \( \sigma \) represents the population standard deviation, and \( n \) is the sample size.The standard error decreases as the sample size increases since you are essentially dividing the population standard deviation by a larger number. A smaller SE results in a tighter confidence interval, indicating a higher degree of confidence in the sample mean as an estimate of the population mean.
In the context of the exercise, as the sample size increased from 49 to 225, the standard error decreased from 1.714 to 0.8, reflecting more precision in the estimate.
Z-Score
The z-score is a statistical measurement that indicates how many standard deviations an element is from the mean. For confidence intervals, the z-score corresponds to the desired level of confidence. For instance, a 95% confidence interval commonly uses a z-score of 1.96. This z-score signifies that the interval is expected to capture the true population mean 95% of the time if the same sampling method is repeated indefinitely. Using a different z-score will adjust the width of the confidence interval. For example, a 90% confidence interval would use a lower z-score (around 1.645), resulting in a narrower interval. Conversely, a 99% confidence interval would need a higher z-score (around 2.576), making the interval wider.
The choice of z-score is crucial because it reflects the confidence level, showing the trade-off between certainty and precision in estimates. In this exercise, a z-score of 1.96 was used consistently to maintain a 95% confidence level across different sample sizes.

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Most popular questions from this chapter

What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p,\) how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.05 from \(p ?\) (b) The Statistical Abstract of the United States, 1 12th edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let \(p\) be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate \(p,\) how large a sample is necessary to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\) (b) In a preliminary random sample of 38 small businesses, it was found that six had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\)

In a marketing survey, a random sample, of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief explanation of the meaning of the interval. (c) As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Thirty small communities in Connecticut (population near 10,000 each) gave an average of \(\bar{x}=138.5\) reported cases of larceny per year. Assume that \(\sigma\) is known to be 42.6 cases per year (Reference: Crime in the United States, Federal Bureau of Investigation). (a) Find a \(90 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (b) Find a \(95 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (c) Find a \(99 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? (c) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

Consider two independent binomial experiments. In the first one, 40 trials had 15 successes. In the second one, 60 trials had 6 successes. (a) Is it appropriate to use a normal distribution to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (c) IBased on the confidence interval you computed, can you be \(95 \%\) confident that \(p_{1}\) is more than \(p_{2} ?\) Explain.

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