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According to an IRS study, it takes an average of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. A consumer watchdog agency selects a random sample of 40 taxpayers and finds the standard deviation of the time to prepare, copy, and electronically file form 1040 is 80 minutes. a. What assumption or assumptions do you need to make about the shape of the population? b. What is the standard error of the mean in this example? c. What is the likelihood the sample mean is greater than 320 minutes? d. What is the likelihood the sample mean is between 320 and 350 minutes? e. What is the likelihood the sample mean is greater than 350 minutes?

Short Answer

Expert verified
a normal distribution; 12.65; 0.7852; 0.7281; 0.0571.

Step by step solution

01

Assumptions about Population Shape

To apply the Central Limit Theorem, we need to assume the distribution of the population is normal, or that our sample size is sufficiently large to assume the sampling distribution of the sample mean is approximately normal. In this case, a sample of 40 is generally considered large enough, so we assume a normal distribution of the sample mean.
02

Calculate the Standard Error of the Mean

The Standard Error of the Mean (SEM) can be calculated using the formula: \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 80 \) is the population standard deviation and \( n = 40 \) is the sample size. Substitute the values: \[ \text{SEM} = \frac{80}{\sqrt{40}} \approx \frac{80}{6.32} \approx 12.65 \]
03

Find Probability Sample Mean > 320 Minutes

To find the probability that the sample mean is greater than 320 minutes, first calculate the z-score. Use the formula: \[ z = \frac{\bar{x} - \mu}{\text{SEM}} \]where \( \bar{x} = 320 \), \( \mu = 330 \), and \( \text{SEM} \approx 12.65 \).\[ z = \frac{320 - 330}{12.65} \approx \frac{-10}{12.65} \approx -0.79 \]Check the Z-table for \( z = -0.79 \) to find the probability, which is approximately 0.2148. So, \( P(\bar{x} > 320) = 1 - 0.2148 = 0.7852 \).
04

Find Probability 320 < Sample Mean < 350

Calculate z-scores for both 320 and 350.For 320:\[ z = \frac{320 - 330}{12.65} \approx -0.79 \] For 350:\[ z = \frac{350 - 330}{12.65} \approx \frac{20}{12.65} \approx 1.58 \]Check the Z-table: - Probability for 320 (\( z = -0.79 \)) is 0.2148.- Probability for 350 (\( z = 1.58 \)) is 0.9429. Thus, \( P(320 < \bar{x} < 350) = 0.9429 - 0.2148 = 0.7281 \).
05

Find Probability Sample Mean > 350 Minutes

Using the z-score from the previous step for 350, which is \( z = 1.58 \), the probability from the Z-table is 0.9429. So, \( P(\bar{x} > 350) = 1 - 0.9429 = 0.0571 \).

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a powerful statistical concept that helps make inferences about population parameters using sample data. It states that, regardless of the population's distribution, the distribution of the sample mean will approach a normal distribution as the sample size becomes larger. This means that the larger the sample size, the more the sampling distribution of the mean tends to form a bell-shaped curve similar to a normal distribution.

In practical terms, the CLT allows statisticians to make predictions and calculate probabilities even if the original population distribution is not normally distributed, as long as the sample size is sufficiently large. In our scenario, examining the time to process tax forms, we used a sample size of 40, which is typically considered large enough under the CLT to assume a normal distribution of the sample mean.

Important points about the CLT include:
  • The sample size should ideally be 30 or more for the CLT to apply effectively.
  • The CLT facilitates approximation of the sampling distribution of the mean even when the population distribution is unknown.
  • This concept is vital because it provides a foundation for statistical inference, including hypothesis testing and confidence interval construction.
Standard Error of the Mean
The Standard Error of the Mean (SEM) is an essential statistical metric that measures the amount of variability or dispersion in a sample mean relative to the true population mean. SEM provides insight into how much discrepancy we might expect between sample means and the population mean across different samples.

The formula for SEM is given by:
\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]
where \(\sigma\) represents the population standard deviation, and \(n\) is the sample size. In our tax form example, \(\sigma = 80\) minutes and \(n = 40\). Thus, the SEM is calculated as follows:
\[ \text{SEM} = \frac{80}{\sqrt{40}} \approx 12.65 \]

Key points about SEM include:
  • SEM decreases as the sample size \(n\) increases, meaning larger samples provide more precise estimates of the population mean.
  • SEM is fundamental in constructing confidence intervals, as it helps define the breadth or precision of the interval.
  • A smaller SEM translates to sample means being closer to the true population mean, indicating more reliability in predictions.
Sampling Distribution
The concept of a Sampling Distribution is foundational in understanding how sample statistics relate to population parameters. A sampling distribution is essentially a probability distribution of a statistic obtained from a large number of samples drawn from a population.

The sampling distribution of the sample mean is particularly important since it provides a basis for estimating the population mean. With the Central Limit Theorem, we know this distribution will be approximately normal for sufficiently large sample sizes, regardless of the population distribution.

In our exercise with taxpayer preparation times, the sampling distribution of the sample mean allows us to predict how sample means vary around the population mean of 330 minutes. This enables us to estimate probabilities and draw conclusions about the population. For instance:
  • The sampling distribution helps calculate how likely a particular sample mean will occur.
  • It enables the computation of probabilities using z-scores, which compare sample means to the expected population mean under normal distribution assumptions.
  • Understanding sampling distribution is crucial for making valid inferences and testing hypotheses about a population based on sample data.
Normal Distribution
The Normal Distribution is a critical concept in statistics, characterized by its symmetrical, bell-shaped curve. It describes how data values are dispersed around the mean, with most of the observations clustering around the center. The importance of normal distribution is immense, particularly because it is the underlying assumption in many statistical tests and methodologies.

Key features of normal distribution include:
  • The mean, median, and mode are all the same and located at the center.
  • Approximately 68% of data falls within one standard deviation from the mean, about 95% within two, and nearly all within three standard deviations.
In our tax preparation example, we assumed that the sampling distribution of the sample mean follows a normal distribution, which facilitated calculating probabilities concerning sample means. For example, determining the likelihood that the average time is more than or less than certain values involves translating these figures into z-scores and using the properties of the normal distribution.

The normal distribution's predictability and mathematical tractability make it essential in assessing how sample data reflects the underlying population.

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Most popular questions from this chapter

At the downtown office of First National Bank there are five tellers. Last week the tellers made the following number of errors each: \(2,3,5,3,\) and \(5 .\) a. How many different samples of 2 tellers are possible? b. List all possible samples of size 2 and compute the mean of each. c. Compute the mean of the sample means and compare it to the population mean.

The Crossett Trucking Company claims that the mean weight of their delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weiqhed. Within what limits will 95 percent of the sample means occur?

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