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The distribution of the number of eggs laid by a certain species of hen during their breeding period has a mean of 35 eggs with a standard deviation of \(18.2 .\) Suppose a group of researchers randomly samples 45 hens of this species, counts the number of eggs laid during their breeding period, and records the sample mean. They repeat this 1,000 times, and build a distribution of sample means. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution and state the appropriate term used to refer to this value. (d) Suppose the researchers' budget is reduced and they are only able to collect random samples of 10 hens. The sample mean of the number of eggs is recorded, and we repeat this 1,000 times, and build a new distribution of sample means. How will the variability of this new distribution compare to the variability of the original distribution?

Short Answer

Expert verified
(a) Sampling distribution of the sample mean. (b) Symmetric; it's approximately normal due to the Central Limit Theorem. (c) Standard error of approximately 2.715. (d) Variability increases with smaller sample size.

Step by step solution

01

Identify the Distribution

The distribution formed by collecting sample means is known as the **sampling distribution (of the sample mean)**.
02

Determine the Shape of the Distribution

According to the Central Limit Theorem, the shape of the sampling distribution of the sample mean will be approximately normal (symmetric) regardless of the shape of the original population distribution, given a sufficiently large sample size (in this case 45).
03

Calculate the Variability

The variability of the sampling distribution is calculated using the formula for the **standard error** (SE) of the mean: \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. Here, \( \sigma = 18.2 \) and \( n = 45 \). So, \( SE = \frac{18.2}{\sqrt{45}} \approx 2.715 \).
04

Compare Variability With a Smaller Sample Size

The variability (standard error) of the sampling distribution increases as the sample size decreases. When the sample size reduces from 45 to 10, the new standard error will be larger. It will be calculated as \( SE = \frac{18.2}{\sqrt{10}} \), which is larger than when \( n = 45 \).

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Key Concepts

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

Sampling Distribution
When researchers repeatedly take a sample and calculate a statistic, such as the mean, from each, the collection of these statistics forms what is known as a sampling distribution. In the context of the exercise, the researchers took 1,000 samples of hen egg counts and calculated the mean for each.

The resulting distribution of these sample means is what we refer to as a sampling distribution.
This concept is crucial in statistics as it helps us make inferences about a population based on a sample. The Central Limit Theorem tells us that, regardless of the shape of the underlying population, the sampling distribution will tend toward a normal distribution as the sample size increases. This characteristic allows statisticians to apply the principles of normal distributions to calculate probabilities and make predictions confidently.
Standard Error
Standard Error (SE) provides a measure of how much sample means differ from the actual population mean. It offers insight into the variability of the sample means.

In mathematical terms, the standard error of the mean is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This formula highlights two important points:
  • A larger sample size \( n \) results in a smaller SE, indicating less variability among sample means.
  • A larger population standard deviation \( \sigma \) will increase the SE, implying greater variability.
Using the given example where \( \sigma = 18.2 \) and \( n = 45 \), the SE was found to be approximately 2.715. This value tells us that the sample means will typically deviate from the true population mean by about 2.715 eggs.
Normal Distribution
Normal distribution is a foundational concept in statistics, known for its bell-shaped curve that is symmetrical around the mean. The Central Limit Theorem reassures us that the sampling distribution of the sample mean will approximate a normal distribution as long as the sample size is sufficiently large, usually \( n \geq 30 \).

This happens even if the original data set (such as the egg counts for individual hens) is not normally distributed.

In the scenario given, with a sample size of 45 hens (which is over 30), the distribution of sample means is expected to be normal. This normality means it is predictable and easier to analyze. It also allows researchers to apply various statistical methods that assume normality, helping in hypothesis testing, confidence interval construction, and other inferential statistics activities.
Sample Size Effect
The size of the sample drawn for a study significantly affects the variability of the sampling distribution. A larger sample size tends to produce a distribution with less variability, meaning the sample means are closer together.

The effect of sample size is embedded in the formula for standard error: \( SE = \frac{\sigma}{\sqrt{n}} \). Here, an increase in n reduces SE, because the larger denominator makes the whole fraction smaller. This is why when researchers cut their sample size from 45 to 10, as described in the exercise, they observed an increase in variability.

The standard error became larger, indicating that means are spread further from the population mean in smaller samples. This reinforces the idea that a larger sample size not only enhances precision but also gives us more reliable and consistent sample means.

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Most popular questions from this chapter

A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $$\$ 100$$. He wants to collect data such that he can get a margin of error of no more than $$\$ 10$$ at a \(95 \%\) confidence level. How large of a sample should he collect?

For a given confidence level, \(t_{d f}^{\star}\) is larger than \(z^{\star}\). Explain how \(t_{d f}^{*}\) being slightly larger than \(z^{*}\) affects the width of the confidence interval.

An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical \(t\) -value \(\left(\mathrm{t}^{\star}\right)\) for the given sample size and confidence level. (a) \(n=6, \mathrm{CL}=90 \%\) (b) \(n=21, \mathrm{CL}=98 \%\) (c) \(n=29, \mathrm{CL}=95 \%\) (d) \(n=12, \mathrm{CL}=99 \%\)

Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of \(124.32 \mu \mathrm{g} / 1\) and a \(\mathrm{SD}\) of \(37.74 \mu \mathrm{g} / \mathrm{l} ;\) a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of \(35 \mu \mathrm{g} / \mathrm{l} .8\) (a) Write down the hypotheses that would be appropriate for testing if the police officers appear to have been exposed to a different concentration of lead. (b) Explicitly state and check all conditions necessary for inference on these data. (c) Regardless of your answers in part (b), test the hypothesis that the downtown police officers have a higher lead exposure than the group in the previous study. Interpret your results in context.

In Exercise 7.24, we discussed diamond prices (standardized by weight) for diamonds with weights 0. 99 carats and 1 carat. See the table for summary statistics, and then construct a \(95 \%\) confidence interval for the average difference between the standardized prices of 0.99 and 1 carat diamonds. You may assume the conditions for inference are met. $$\begin{array}{lcc} \hline & 0.99 \text { carats } & \text { 1 carat } \\ \hline \text { Mean } & \$ 44.51 & \$ 56.81 \\ \text { SD } & \$ 13.32 & \$ 16.13 \\ \text { n } & 23 & 23 \\ \hline \end{array}$$

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