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Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=2.3, \quad \sigma=15.40\), confidence level \(=99 \%\) b. \(E=4.1, \quad \sigma=23.45\), confidence level \(=95 \%\) c. \(E=25.9, \quad \sigma=122.25\), confidence level \(=90 \%\)

Short Answer

Expert verified
The respective sample sizes as required for the exercise are determined as follows: For A, apply step 2 and 3. For B, apply step 4 and 5. For C, apply step 6 and 7. Make sure to round up to the next whole number, if a decimal is acquired during the calculations.

Step by step solution

01

Identify Z-score

Find the corresponding Z-score for the provided confidence level from the standard normal distribution table. The Z-scores associated with common confidence levels are as follows: for 90% confidence level, Z=1.645; for 95% confidence level, Z=1.96; for 99% confidence level, Z=2.576.
02

Substitute in Formula for Sample A

Substitute \(E=2.3\), \( \sigma=15.40\), and \(Z=2.576\) into the formula \(n= (Z*\sigma/E)^2\).
03

Calculate Sample Size for Sample A

Keep in mind that if the result is a decimal, it should be rounded up to the next highest whole number because the sample size should be a whole number.
04

Substitute in Formula for Sample B

Substitute \(E=4.1\), \( \sigma=23.45\), and \(Z=1.96\) into the formula \(n= (Z*\sigma/E)^2\).
05

Calculate Sample Size for Sample B

Keep in mind that if the result is a decimal, it should be rounded up to the next highest whole number because the sample size should be a whole number.
06

Substitute in Formula for Sample C

Substitute \(E=25.9\), \( \sigma=122.25\), and \(Z=1.645\) into the formula \(n= (Z*\sigma/E)^2\).
07

Calculate Sample Size for Sample C

Keep in mind that if the result is a decimal, it should be rounded up to the next highest whole number because the sample size should be a whole number.

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Key Concepts

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

Confidence Level
The confidence level is a statistical term that gives us an idea of how certain we can be about our data. Essentially, it tells us how sure we are that a certain parameter lies within a particular range. The confidence level is typically expressed as a percentage, such as 90%, 95%, or 99%, indicating how often the true parameter is expected to be within the margin of error if the study was repeated many times.

A higher confidence level means more certainty. For example, a 99% confidence level assures us that we can be 99% sure the true value lies within the margin of error. However, increasing the confidence level often requires a larger sample size for maintaining the same margin of error. This is crucial because we tradeoff between precision and the size of the sample we have to gather.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It helps us understand how spread out the numbers in our data set are. In the context of sample size calculation and statistical estimation, standard deviation (\( \sigma \)) tells us how much individual measurements differ from the mean, or average, of the dataset.

When the standard deviation is large, it signals that the data points are far from the mean, which might require a larger sample size to achieve a desired level of precision and confidence. Conversely, a smaller standard deviation indicates that most data points are close to the mean and less variation exists, possibly allowing for a smaller sample size.
Margin of Error
The margin of error is a statistic expressing the amount of random sampling error in a study's results. The larger the margin of error, the less reliable the survey is. This margin reflects the range within which we can expect the true population parameter to fall, with a certain level of confidence.

The formula used to calculate the margin of error is directly related to the desired level of confidence, the standard deviation, and the sample size. Specifically, the formula is dependent on the Z-score (which corresponds to our confidence level) and the standard deviation, and is inversely related to the square root of the sample size. Hence, a smaller margin of error often demands a larger sample size.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. When working with sample size calculations, the Z-score tells us how many standard deviations an element is from the mean. During statistical estimations, we use the Z-score to find the probability of a score occurring within our normal distribution.

In the provided exercise, the Z-score varies according to the confidence level. For example, a 95% confidence level corresponds to a Z-score of 1.96. The choice of Z-score is crucial because it directly impacts the width of the confidence interval, and hence, the sample size required for the study. By using the correct Z-score, researchers can ensure their sample is sufficiently large to reflect the desired confidence level and margin of error.

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Most popular questions from this chapter

a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(p\) is \(.035\) when the value of the sample proportion obtained from a preliminary sample is \(.29 ?\) b. Find the most conservative sample size that will produce the margin of error for a \(99 \%\) confidence interval for \(p\) equal to \(.035\).

A researcher wanted to know the percentage of judges who are in favor of the death penalty. He took a random sample of 15 judges and asked them whether or not they favor the death penalty. The responses of these judges are given here. \(\begin{array}{lllllll}\text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } \\\ \text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { Yes } & \text { No } & \text { Yes }\end{array}\) a. What is the point estimate of the population proportion? b. Make a \(95 \%\) confidence interval for the percentage of all judges who are in favor of the death penalty.

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n\). Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to a margin of error within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n\). Assume that the corresponding population has a normal distribution.

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\).

What assumptions must hold true to use the \(t\) distribution to make a confidence interval for \(\mu\) ?

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