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Chapter 7: Problem 14

For each of the following three sample sizes, construct the \(95 \%\) confidence interval. Use a sample proportion of \(0.40\) throughout. What happens to interval width as sample size increases? Why? $$ \begin{aligned} P_{s} &=0.40 \\ \text { Sample A: } N &=100 \\ \text { Sample B: } N &=1000 \\ \text { Sample C: } N &=10,000 \end{aligned} $$

Short Answer

Expert verified
The confidence interval width decreases as sample size increases because the standard error decreases with larger sample sizes.

Step by step solution

01

Identify the given parameters

The sample proportion, denoted as \(P_s\), is given as 0.40. The confidence level is 95%, which corresponds to a Z-score (\

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Key Concepts

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

Sample Size
The term 'sample size' refers to the number of observations or data points collected in a study. It is denoted by the letter N. The importance of sample size in statistics cannot be overstated, as it directly affects the accuracy and reliability of the results. A larger sample size provides more information and tends to produce more reliable estimates.
In the context of our exercise, we have three different sample sizes: N = 100, N = 1000, and N = 10,000.
  • Sample A: N = 100
  • Sample B: N = 1000
  • Sample C: N = 10,000
As the sample size increases from 100 to 10,000, the confidence intervals will become narrower because larger sample sizes reduce the standard error.
Proportion
The term 'proportion' represents the fraction of the total that possesses a particular characteristic of interest. It is often denoted as P or \(P_s\). In our problem, the sample proportion (\(P_s\)) is given as 0.40.

This means in our sample, 40% of the observations possess the trait we are analyzing. Calculating the confidence interval around this proportion will allow us to understand the range within which the true population proportion is likely to fall.

  • Proportion = Number of favorable outcomes / Total number of outcomes
  • \(P_s = 0.40\)
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations from the mean. In the context of confidence intervals, the Z-score is used to determine the margin of error.

For a 95% confidence level, the Z-score is approximately 1.96. This value comes from standard normal distribution tables and signifies that 95% of the data falls within 1.96 standard deviations of the mean.

  • Z-score for 95% confidence level = 1.96
Calculating the Z-score helps us to adjust our interval to the desired confidence level, ensuring that our interval estimate is accurate.
Interval Width
The interval width in a confidence interval is the range between the lower and upper bounds. It is influenced by both the sample size and the standard error. A narrower interval width indicates more precise estimates.

As shown in the exercise, as the sample size increases from 100 to 10,000, the interval width decreases. This happens because the standard error, which is inversely proportional to the square root of the sample size, gets smaller.

The interval width is given by the formula: \[ \text{Interval Width} = 2 \times (Z \times \text{Standard Error}) \]
  • Sample A: Wider interval
  • Sample B: Narrower interval
  • Sample C: Narrowest interval
Increasing the sample size improves the precision of our confidence interval, thereby reducing the interval width.

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Most popular questions from this chapter

A random sample of 429 college students was interviewed about a number of matters. a. They reported that they had spent an average of \(\$ 345.23\) on textbooks during the previous semester. If the sample standard deviation for these data is \(\$ 15.78\), construct an estimate of the population mean at the \(99 \%\) level. b. They also reported that they had visited the health clinic an average of \(1.5\) times a semester. If the sample standard deviation is \(0.3\), construct an estimate of the population mean at the \(99 \%\) level. c. On the average, the sample had missed \(2.8\) days of classes per semester because of illness. If the sample standard deviation is \(1.0\), construct an estimate of the population mean at the \(99 \%\) level. d. On the average, the sample had missed \(3.5\) days of classes per semester for reasons other than illness. If the sample standard deviation is \(1.5, \mathrm{con}-\) struct an estimate of the population mean at the \(99 \%\) level.

A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to estimate the population mean. Use the \(90 \%\) level. a. An average of \(2.3\) people resides in each household. Standard deviation is \(0.35\). b. There was an average of \(2.1\) television sets \((s=0.10)\) and \(0.78\) telephones \((s=0.55)\) per household. c. The households averaged \(6.0\) hours of television viewing per day \((s=3.0)\)

A random sample of } 260 \text { workers in a high-rise }\end{array}\( office building revealed that \)30 \%\( were very satisfied with the quality of elevator service. At the \)99 \%$ level, what is your estimate of the population value?

You have developed a series of questions to measure job satisfaction of bus drivers in New York City. A random sample of 100 drivers has an average score of \(10.6\), with a standard deviation of \(2.8\). What is your estimate of the average job satisfaction score for the population as a whole? Use the \(95 \%\) confidence level.

For each of the following sets of sample outcomes, construct the \(99 \%\) confidence interval for estimating \(P_{u}\). a. \(\begin{aligned} P_{s} &=0.40 \\ N &=548 \end{aligned}\) b. \(\begin{aligned} P_{s} &=0.37 \\ N &=522 \end{aligned}\) c. \(\begin{aligned} P_{s} &=0.79 \\ N &=121 \end{aligned}\) d. \(\begin{aligned} P_{s} &=0.14 \\ N &=100 \end{aligned}\) e. \(\begin{aligned} P_{s} &=0.43 \\ N &=1049 \end{aligned}\) f. \(\begin{aligned} P_{s} &=0.63 \\ N &=300 \end{aligned}\)
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