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a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\).

Short Answer

Expert verified
a. The required sample size for a margin of error of .045 when the preliminary sample proportion is .53 is 813. b. The most conservative sample size for a margin of error of .045 is 853.

Step by step solution

01

Identify given values

From the exercise, we have the following values: preliminary sample proportion \(p = .53\), margin of error \(E = .045\), and confidence level \(CL = 98 \% = 0.98\). For a two-tailed test with a confidence level of \(98\% \), the \(Z_{\alpha/2}\) value (the z-score that accumulates \((1-CL)/2 = 0.01\) on each tail) can be looked up in the standard normal table or can be calculated using a statistical software or calculator. The resulting \(Z_{\alpha/2}\) is approximately 2.33.
02

Calculate sample size for part a

Plug the known values into the sample size formula: \(n = \frac{(Z_{\alpha/2})^2 \times p \times (1-p)}{E^2}\). Hence \(n = \frac{ (2.33)^2 \times 0.53 \times (1-0.53) }{ (0.045)^2 } = 812.70\). Since we cannot have a decimal of a person, round up to the nearest whole number. Thus, the required sample size for part a is 813.
03

Calculate most conservative sample size for part b

The most conservative estimate of the sample size is achieved when \(p = 0.5\) (since the product \(p(1 - p)\) is maximum at this point). So to get the most conservative sample size, plug \(p = 0.5\) into the sample size formula. Then, \(n = \frac{ (2.33)^2 \times 0.5 \times (1-0.5) }{ (0.045)^2 } = 852.93\). Again, round up to the next whole number. Hence, the most conservative sample size for part b is 853.

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Key Concepts

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

Confidence Interval
A confidence interval is a statistical tool used to estimate the range in which a population parameter lies based on sample data. For example, in our exercise, we're focused on estimating the population proportion \( p \).

This interval provides a range that we are fairly confident includes the true population parameter. The confidence level, here \(98\%\), indicates the degree of certainty that the interval contains the population parameter. A \(98\%\) confidence level means that if we were to take 100 different samples and compute a confidence interval for each, we'd expect about 98 of them to contain the true value of the parameter.

Confidence intervals are valuable because they provide more information than a simple point estimate, like a sample mean or proportion. They reflect the uncertainty and variability that come with taking a sample instead of examining the entire population.
Margin of Error
The margin of error represents the range within which the true population parameter is expected to lie from the estimated value. In our exercise, the margin of error \( E \) is \(0.045\).

This means the sample proportion could be higher or lower by 4.5% from the true population proportion. The margin of error depends on the confidence level and the variability within the population, which is often estimated using preliminary sample data.

The smaller the margin of error, the closer our sample estimate is expected to be to the actual population parameter. However, to achieve a smaller margin of error, we may need a larger sample size, as seen in this exercise.
Z-Score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations from the mean. For confidence intervals, the z-score (\(Z_{\alpha/2}\)) is crucial because it adjusts the width of the confidence interval based on the desired confidence level.

For a 98% confidence interval, the z-score \(Z_{\alpha/2}\) is approximately 2.33. This value can be found using a standard normal distribution table. The higher the confidence level, the larger the z-score, and subsequently, the wider the confidence interval, which affects the margin of error.

In summary, a higher z-score indicates a broader interval, thus providing more certainty but requiring a larger sample size to maintain the same margin of error.
Statistical Formulas
Statistical formulas are key to determining the optimal sample size. In our exercise, we use the formula for determining sample size:
\[ n = \frac{(Z_{\alpha/2})^2 \times p \times (1-p)}{E^2}\]

Where
  • \(n\) is the required sample size
  • \(Z_{\alpha/2}\) is the z-score associated with the confidence level
  • \(p\) is the sample proportion
  • \(E\) is the margin of error

For the most conservative sample size, \(p\) is set to 0.5, as this value maximizes the product \(p(1-p)\), ensuring sufficient sample size regardless of the actual proportion.

This formula helps balance the precision of the estimate against the effort needed to obtain a sufficiently large sample.

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Most popular questions from this chapter

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110\) and \(\hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)

What is the point estimator of the population proportion, \(p\) ?

You are working for a bank. The bank manager wants to know the mean waiting time for all customers who visit this bank. She has asked you to estimate this mean by taking a sample. Briefly explain how you will conduct this study. Collect data on the waiting times for 45 customers who visit a bank. Then estimate the population mean. Choose your own confidence level

Because of inadequate public school budgets and lack of money available to teachers for classroom materials, many teachers often use their own money to buy materials used in the classrooms. A random sample of 100 public school teachers selected from an eastern state showed that they spent an average of \(\$ 273\) on such materials during the 2009 school year. The population standard deviation was \$60. a. What is the point estimate of the mean of such expenses incurred during the 2009 school year by all public school teachers in this state? b. Make a \(95 \%\) confidence interval for the corresponding population mean.

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

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