Problem 18 - Basic Statistics for Business and Economics with Student

Question

According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. a. What is the standard error of the mean in this example? b. What is the likelihood the sample mean is greater than 320 minutes? c. What is the likelihood the sample mean is between 320 and 350 minutes? d. What is the likelihood the sample mean is greater than 350 minutes?

Step-by-Step Solution

Answer

a. SE = 12.65; b. P > 320 = 0.7852; c. P(320 < X < 350) = 0.7281; d. P > 350 = 0.0571.

Step 1: Compute the Standard Error of the Mean

The standard error (SE) 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.- Given \( \sigma = 80 \) minutes and \( n = 40 \),- \[ SE = \frac{80}{\sqrt{40}} \approx 12.65 \]Thus, the standard error of the mean is approximately 12.65 minutes.

Step 2: Find Probability of Sample Mean Greater Than 320 Minutes

First, calculate the z-score for 320 minutes using the formula:\[ z = \frac{X - \mu}{SE} \]where \( X \) is the sample mean, \( \mu = 330 \), and \( SE = 12.65 \).- For 320 minutes:- \[ z = \frac{320 - 330}{12.65} \approx -0.79 \]Next, find the probability using the standard normal distribution table or calculator. The probability of \( z > -0.79 \) is approximately 0.7852. Therefore, the likelihood the sample mean is greater than 320 minutes is 0.7852.

Step 3: Calculate Probability of Sample Mean Between 320 and 350 Minutes

Calculate the z-score for 350 minutes:- \[ z = \frac{350 - 330}{12.65} \approx 1.58 \]Find the probabilities:- Probability of \( z < 1.58 \) is approximately 0.9429.- Probability of \( z < -0.79 \) is approximately 0.2148.- The probability that the sample mean is between 320 and 350 minutes is the difference:- \[ P(320 < X < 350) = P(z < 1.58) - P(z < -0.79) \approx 0.9429 - 0.2148 = 0.7281 \]

Step 4: Determine Probability of Sample Mean Greater Than 350 Minutes

Use the z-score for 350 minutes calculated previously:- \[ z = 1.58 \]Find the probability of \( z > 1.58 \), which is:- \[ 1 - P(z < 1.58) \approx 1 - 0.9429 = 0.0571 \]Thus, the likelihood the sample mean is greater than 350 minutes is approximately 0.0571.

Key Concepts

Normal Distribution

The concept of normal distribution is fundamental when dealing with data that tends to cluster around a mean. In the exercise given, the preparation times for taxpayers follow a normal distribution, meaning they adhere to a pattern often observed in natural phenomena. Normal distributions are symmetric, forming a bell-shaped curve, and are defined by two parameters: the mean and the standard deviation.

• The mean is the average value, which in this case is 330 minutes.
• The standard deviation, 80 minutes, measures the amount of variation or dispersion from the mean.

In a normal distribution, most data points will fall within three standard deviations of the mean. In real-world scenarios, understanding this distribution allows statisticians to make inferences and predictions based on data.

Z-Score Calculation

When analyzing how likely it is for a sample mean to differ from the population mean, the z-score is a crucial tool. A z-score tells us how many standard deviations a point is from the mean. To find the z-score, we use the formula: \[ z = \frac{X - \mu}{SE} \]where:

• \( X \) is the value of interest, the sample mean we want to investigate.
• \( \mu \) is the population mean.
• \( SE \) is the standard error of the mean.

For example, in the problem, calculating the z-score for a sample mean of 320 minutes helps us determine how far this sample mean is from the average taxpayer's preparation time. This calculation serves as a basis for determining probabilities using the normal distribution.

Probability Calculation

With z-scores calculated, we can now determine probabilities, or the likelihood of observing a sample mean in a specified range. Probabilities are found by consulting a standard normal distribution table or using a calculator. These values represent the proportion of samples expected to fall within certain z-score brackets.
By solving the z-scores for 320 minutes and 350 minutes, we found the values necessary to determine two specific probabilities:

• The probability that the sample mean exceeds 320 minutes.
• The probability that the sample mean is between 320 and 350 minutes.

Using these data points, we can predict and interpret the behavior of the dataset as a whole, which offers valuable insights into the distribution of taxpayer preparation times.

Sampling Distribution

Sampling distribution is a critical concept, especially when making inferences about a population based on a sample. It refers to the probability distribution of a given statistic—such as the mean—based on a random sample. This is essential because the sample mean varies in a known way, allowing us to calculate probabilities about where it is likely to fall.
For instance, by understanding this concept, we can compute the standard error of the mean. The formula for standard error, \( SE = \frac{\sigma}{\sqrt{n}} \), allows us to measure how much the sample mean of different samples will vary. In our scenario, with the population standard deviation of 80 minutes and a sample size of 40, the standard error was approximately 12.65 minutes.
Through these calculations and concept comprehension, it becomes clear how sampling distributions serve as a bridge, connecting sample data to population parameters, thereby supporting decisive and informed statistical conclusions.