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For a population data set, \(\sigma=14.50\). a. What should the sample size be for a \(98 \%\) confidence interval for \(\mu\) to have a margin of error of estimate equal to \(5.50 ?\) b. What should the sample size be for a \(95 \%\) confidence interval for \(\mu\) to have a margin of error of estimate equal to \(4.25 ?\)

Short Answer

Expert verified
To achieve a margin of error of \(5.50\) with a \(98\%\) confidence interval, a sample size of 30 is needed. To achieve a margin of error of \(4.25\) with a \(95\%\) confidence interval, a sample size of 61 is needed.

Step by step solution

01

Identify the given values for part a

The population standard deviation \(\sigma=14.50\), the desired margin of error \(E=5.50\), and the z-value for a \(98\%\) confidence interval (which corresponds to \(Z=2.33\)).
02

Apply formula to calculate sample size for part a

Apply the formula \(\frac {(Z^2 * σ^2)}{E^2}\) to calculate the sample size. Substituting the given values into the formula gives \(\frac {(2.33^2 * 14.50^2)}{5.50^2}=29.96\). We always round up to the nearest whole number when calculating sample size, so the sample size needed is 30.
03

Identify the given values for part b

The population standard deviation \(\sigma=14.50\), the desired margin of error \(E=4.25\), and the z-value for a \(95\%\) confidence interval (which corresponds to \(Z=1.96\)).
04

Apply formula to calculate sample size for part b

Again, apply the formula \(\frac {(Z^2 * σ^2)}{E^2}\) to calculate the sample size. Substituting the given values into the formula gives \(\frac {(1.96^2 * 14.50^2)}{4.25^2}=60.95\). Again, rounding up to the nearest whole number, the sample size needed is 61.

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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 indicates the range that the true population parameter—such as the mean—can be expected to fall within, given our sample data and confidence level. It tells us how much we can expect our sample results to differ from the true population mean, purely due to random sampling. If, for instance, we calculate a margin of error of 5.50, this means our estimate for the mean is off by 5.50 units at most.

Importantly, a smaller margin of error indicates more precise results, often requiring a larger sample size. Therefore, understanding how to set and achieve the desired margin of error according to the study's requirements is crucial in statistical analysis.
Confidence Interval
A confidence interval not only tells us about the potential values the population mean might take, but also measures the uncertainty surrounding the sample estimate. It consists of a range of values, calculated from the sample, that likely contains the true population parameter. When we say we want a 98% confidence interval, we mean we want to be 98% sure that this interval captures the true average.

Confidence intervals give us valuable insights by providing both the estimate, like the mean, and a sense of the estimate's reliability. They are calculated using the formula: \[ \left( \text{Sample Mean} - Z\frac{\sigma}{\sqrt{n}}, \ \text{Sample Mean} + Z\frac{\sigma}{\sqrt{n}} \right) \]where \(Z\) is the critical value linked to the confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
Population Standard Deviation
The population standard deviation, denoted as \(\sigma\), represents the dispersion of the population data points around the mean. In simple terms, it tells us how much the individual data points in a population deviate from the average value. A larger \(\sigma\) signifies a more spread-out population, suggesting more variability in the data.

When calculating sample sizes, knowing \(\sigma\) is crucial because it helps us determine how large a sample we might need to achieve a desired level of precision (or margin of error) in our confidence intervals. In situations where the population standard deviation is high, you would typically need a larger sample size than if \(\sigma\) were small.
Z-value
In the context of sample size calculation, the "Z-value" or "Z-score" is a critical value derived from the standard normal distribution. It is associated with your chosen confidence level. For a common confidence interval:
  • A 95% confidence interval corresponds to a Z-value of 1.96.
  • A 98% confidence interval corresponds to a Z-value of 2.33.
Each Z-value represents the number of standard deviations a point can be from the population mean while still falling within the desired confidence level.

In sample size formulas, the Z-value is multiplied by the population standard deviation and used in conjunction with the margin of error, ensuring our interval is wide enough to confidently include the population mean.
Rounding up sample size
The process of rounding up the sample size ensures that our estimates remain conservative and precise. In statistical analysis, any calculated sample size with decimal points is increased to the next whole number. This practice stems from the need to meet or exceed the desired confidence level and margin of error.

For example, when computing a sample size and arriving at 29.96, you would round to 30 to ensure your sample is sufficiently large. By rounding up, you make sure the interval is wide enough to encompass the true population parameter confidently, based on your calculated precision.

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Most popular questions from this chapter

A sample of 18 observations taken from a normally distributed population produced the following data. \(\begin{array}{lllllllll}28.4 & 27.3 & 25.5 & 25.5 & 31.1 & 23.0 & 26.3 & 24.6 & 28.4\end{array}\) \(\begin{array}{llllllll}37.2 & 23.9 & 28.7 & 27.9 & 25.1 & 27.2 & 25.3\end{array}\) \(\begin{array}{ll}22.6 & 22.7\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for \(\mu\) in part b?

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

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You want to estimate the proportion of students at your college who hold off- campus (part-time or full-time) jobs. Briefly explain how you will make such an estimate. Collect data from 40 students at your college on whether or not they hold off-campus jobs. Then calculate the proportion of students in this sample who hold off-campus jobs. Using this information, estimate the population proportion. Select your own confidence level.

Salaried workers at a large corporation receive 2 weeks' paid vacation per year. Sixteen randomly selected workers from this corporation were asked whether or not they would be willing to take a \(3 \%\) reduction in their annual salaries in return for 2 additional weeks of paid vacation. The following are the responses of these workers. \(\begin{array}{llllllll}\text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } \\ \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No }\end{array}\) Construct a \(97 \%\) confidence interval for the percentage of all salaried workers at this corporation who would accept a \(3 \%\) pay cut in return for 2 additional weeks of paid vacation.

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