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Chapter 8: Problem 3

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

Short Answer

Expert verified
a: [76.206, 83.794]; b: [77.317, 82.683]; c: Larger sample size results in a narrower confidence interval.

Step by step solution

01

Identify the Known Values for Part a

We have a sample of size \( n = 60 \) with a sample mean \( \bar{x} = 80 \). The population standard deviation \( \sigma = 15 \), and we want a 95% confidence interval.
02

Calculate Standard Error for Part a

The standard error (SE) is calculated as \( SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{60}} \approx 1.936 \).
03

Determine the Z-Score for 95% Confidence

For a 95% confidence interval, the Z-score is approximately 1.96.
04

Compute the Confidence Interval for Part a

The confidence interval is given by: \[ \bar{x} \pm Z \times SE \]. Thus, the interval is \[ 80 \pm 1.96 \times 1.936 \]. Calculate to find: \( 80 \pm 3.794 \), resulting in \([76.206, 83.794]\).
05

Identify the Known Values for Part b

Now with \( n = 120 \), the sample mean \( \bar{x} = 80\), and \( \sigma = 15 \) remains the same.
06

Calculate Standard Error for Part b

The standard error is now \( SE = \frac{15}{\sqrt{120}} \approx 1.369 \).
07

Compute the Confidence Interval for Part b

Using the same Z-score of 1.96, the interval becomes: \[ 80 \pm 1.96 \times 1.369 \]. Calculate to find: \( 80 \pm 2.683 \), resulting in \([77.317, 82.683]\).
08

Discussing the Effect of Larger Sample Size

A larger sample size results in a smaller standard error, which makes the confidence interval narrower, indicating a more precise estimate of the population mean.

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Key Concepts

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

Sample Size Effect
Understanding the effect of sample size is crucial when constructing confidence intervals. When we increase the sample size, the estimates of population parameters tend to be more reliable. Larger samples provide more information about the population, leading to more precise estimates.
  • **Larger Sample Size:** When we increase our sample from 60 to 120 items, the confidence interval becomes narrower. This is because the larger the sample, the more the sample mean is likely to be close to the actual population mean.
  • **Narrower Intervals:** With a sample size of 120, the confidence interval calculation gives us \( [77.317, 82.683]\), compared to a broader interval of \( [76.206, 83.794]\) with 60 samples. A narrower interval implies increased precision.
  • **Reduced Margin of Error:** More data means less variability and erratic swings in the estimate, which results in a reduced margin of error. This reflects that our population estimate is stable.
Hence, larger samples provide finer estimates of the population parameter, yielding narrow confidence intervals and reduced error margins.
Standard Error Calculation
Standard error is a key component when working with confidence intervals. It measures the variability of a sample statistic, such as the sample mean, from the population parameter.
  • **Formula:** The formula for standard error (SE) is given by \( SE = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) represents the population standard deviation and \(n\) is the sample size.
  • **Impact of Sample Size:** When calculating standard error, we see that as the sample size \(n\) increases, the denominator of the formula increases, resulting in a smaller standard error. In our example, increasing the sample size from 60 to 120 decreases the SE from approximately 1.936 to about 1.369.
  • **Precision in Estimates:** A smaller standard error means that the sample mean is a more accurate reflection of the population mean. Less variation in the sample mean enhances the precision of our confidence interval estimates.
In summary, the standard error is inversely related to the sample size, playing a crucial role in defining the reliability of our confidence interval.
Z-Score for Confidence Intervals
The Z-score is a statistic that informs us how many standard deviations an element lies from the mean. In confidence intervals, it determines the width of the interval with a given confidence level.
  • **Use in Confidence Intervals:** For a 95% confidence interval, we often use a Z-score of 1.96. This value is derived from the standard normal distribution, representing the range within which the population mean lies, 95% of the time.
  • **Application in Calculation:** In calculations, we express the confidence interval in the form \(\bar{x} \pm Z \times SE\). In step 4 of our example, the interval is \(80 \pm 1.96 \times 1.936\) for a sample size of 60, and in step 7, with a sample size of 120, it becomes \(80 \pm 1.96 \times 1.369\)
  • **Significance of the Z-score:** The chosen Z-score ensures that the corresponding confidence interval is neither too broad nor too narrow, maintaining the auto balance required for the interval's accuracy and reliability.
Understanding and utilizing the Z-score is essential for constructing and interpreting confidence intervals effectively.

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Most popular questions from this chapter

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \(\$ 5\) a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

Mileage tests are conducted for a particular model of automobile. If a \(98 \%\) confidence interval with a margin of error of 1 mile per gallon is desired, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation is 2.6 miles per gallon.

The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?

According to Thomson Financial, through January \(25,2006,\) the majority of companies reporting profits had beaten estimates (BusinessWeek, February 6, 2006). A sample of 162 companies showed 104 beat estimates, 29 matched estimates, and 29 fell short. a. What is the point estimate of the proportion that fell short of estimates? b. Determine the margin of error and provide a \(95 \%\) confidence interval for the proportion that beat estimates. c. How large a sample is needed if the desired margin of error is \(.05 ?\)

Consumption of alcoholic beverages by young women of drinking age has been increasing in the United Kingdom, the United States, and Europe (The Wall Street Journal, February 15 2006 ). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women. $$\begin{array}{rrrrr}266 & 82 & 199 & 174 & 97 \\\170 & 222 & 115 & 130 & 169 \\\164 & 102 & 113 & 171 & 0 \\\93 & 0 & 93 & 110 & 1300\end{array}$$ Assuming the population is roughly symmetric, construct a \(95 \%\) confidence interval for the mean annual consumption of alcoholic beverages by European young women.
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