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$$ \begin{array}{l}{\text { Christmas Presents How large a sample is needed to }} \\\ {\text { estimate the population mean for the amount of money a }} \\\ {\text { person spends on Christmas presents within } \$ 2 \text { and be } 95 \%} \\ {\text { confident? The standard deviation of the population }} \\\ {\text { is } \$ 7.50 .}\end{array} $$

Short Answer

Expert verified
A sample size of 55 is needed.

Step by step solution

01

Understand the Problem

We need to find the sample size necessary to estimate the population mean expenditure on Christmas presents with a margin of error of $2 and 95% confidence, given that the population standard deviation is $7.50.
02

Identify Key Parameters

Identify the key parameters needed for the calculation: - Margin of error (E) = $2 - Confidence level = 95% - Population standard deviation (σ) = $7.50.
03

Find the Z-Score for 95% Confidence

For a 95% confidence level, use a standard normal distribution table to find the Z-score (Z) that corresponds to the tails of the distribution. The Z-score is approximately 1.96.
04

Apply the Sample Size Formula

Use the formula for sample size: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]Substitute the known values into the formula:\( n = \left( \frac{1.96 \times 7.50}{2} \right)^2 \).
05

Calculate the Sample Size

Calculate the value inside the parentheses first:\( 1.96 \times 7.50 = 14.7 \).Then divide by the margin of error:\( \frac{14.7}{2} = 7.35 \).Finally, square this value:\( n = (7.35)^2 = 54.0225 \).
06

Round Up the Sample Size

Since the sample size must be a whole number, round 54.0225 up to the nearest whole number, so the required sample size is 55.

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Key Concepts

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

Population Mean Estimation
When collecting data, we often aim to estimate characteristics about a whole group or "population" using information from a smaller group known as a "sample." Estimating the population mean involves determining the average value of a characteristic across the entire population. By taking a sample, we can calculate an average from this smaller group, which serves as an estimate of the population mean.

However, simply using a sample average isn't sufficient to claim accurate estimation of the population mean, because it may not perfectly represent the larger population. Therefore, additional statistical techniques are applied to ensure our estimations are as close to the true mean as possible. Using calculations that include the confidence level and margin of error, we optimize our sample size for better precision.
Margin of Error
The margin of error reflects the uncertainty or possible error range in our estimation of the population mean. It quantifies how much the estimated population mean might differ from the sample mean. In the exercise, we wanted the estimation to be within $2 of the actual population mean.

A smaller margin of error means more precision and accuracy in the sample mean—it's closer to what the actual population mean might be. However, reducing the margin inevitably requires larger sample sizes owing to more data needed to achieve higher precision. The balance between obtaining a manageable sample size and achieving desired accuracy is crucial in statistics.
Confidence Level
The confidence level indicates how sure we are that the population mean lies within a specified range around the sample mean. It's expressed as a percentage. In our exercise, we used a 95% confidence level, suggesting we can be 95% certain that the true population mean resides within the margin of error around our expected sample mean.

To determine this range, statisticians use a Z-score associated with the chosen confidence level. A higher confidence level implies more certainty, but it also results in a broader margin of error unless the sample size is increased to maintain the same range. Hence, it's all about finding the right balance between confidence and practicality.
Standard Deviation
Standard deviation tells us about the spread or variability of data in a population or sample. It describes the average distance of each data point from the mean. A larger standard deviation signifies more variability within the data set—meaning data points are spread out over a wider range, while a smaller standard deviation indicates data points are clustered closely around the mean.

In our Christmas presents example, the standard deviation is given as $7.50. This value plays a significant role in calculating sample size because it affects how much the data tends to spread out. When populated into our sample size formula, combined with Z-score and margin of error, it helps determine the number of samples needed to achieve our desired levels of confidence and precision.

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Most popular questions from this chapter

Holiday Gifts A survey of 100 Americans found that \(68 \%\) said they find it hard to buy holiday gifts that convey their true feelings. Find the \(90 \%\) confidence interval of the population proportion.

Undergraduate GPAs It is desired to estimate the mean GPA of each undergraduate class at a large uni- versity. How large a sample is necessary to estimate the GPA within 0.25 at the \(99 \%\) confidence level? The population standard deviation is \(1.2 .\)

High School Graduates Who Take the SAT The national average for the percentage of high school graduates taking the SAT is \(49 \%\), but the state averages vary from a low of \(4 \%\) to a high of \(92 \%\). A random sample of 300 graduating high school seniors was polled across a particular tristate area, and it was found that 195 had taken the SAT. Estimate the true proportion of high school graduates in this region who take the SAT with \(95 \%\) confidence.

Cyber Monday Shopping A survey of \(1000 \mathrm{U.S}\). adults found that \(33 \%\) of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the \(95 \%\) confidence inter- val of the true proportion.

Find the values for each. $$ \begin{array}{l}{\text { a. } t_{\alpha / 2} \text { and } n=18 \text { for the } 99 \% \text { confidence interval for }} \\ {\text { the mean }} \\\ {\text { b. } t_{\alpha / 2} \text { and } n=23 \text { for the } 95 \% \text { confidence interval for }} \\ {\text { the mean }} \\ {\text { c. } t_{\alpha / 2} \text { and } n=15 \text { for the } 98 \% \text { confidence interval for }} \\ {\text { the mean }} \\ {\text { d. } t_{\alpha / 2} \text { and } n=10 \text { for the } 90 \% \text { confidence interval for }} \\\ {\text { the mean }} \\ {\text { e. } t_{\alpha / 2} \text { and } n=20 \text { for the } 95 \% \text { confidence interval for }} \\ {\text { the mean }}\end{array} $$

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