Problem 33 - Basic Statistics for Business and Economics with Student

Question

Recent studies indicate that the typical 50 -year-old woman spends \(\$ 350\) per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of \(\$ 45\) per year. We select a random sample of 40 women. The mean amount spent for those sampled is \(\$ 335 .\) What is the likelihood of finding a sample mean this large or larger from the specified population?

Step-by-Step Solution

Answer

The likelihood is approximately 98.26%.

Step 1: Define the Parameters

We have a normal distribution where the population mean, \( \mu \), is \( 350 \), and the standard deviation, \( \sigma \), is \( 45 \). The sample size, \( n \), is \( 40 \), and the sample mean, \( \bar{x} \), is \( 335 \).

Step 2: Calculate the Standard Error of the Mean

The standard error of the mean (SEM) is calculated using the formula: \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \). Substituting the given values: \[ \text{SEM} = \frac{45}{\sqrt{40}} \approx 7.12. \]

Step 3: Compute the Z-Score

The Z-score for the sample mean is given by the formula: \( Z = \frac{\bar{x} - \mu}{\text{SEM}} \). Substituting the values: \[ Z = \frac{335 - 350}{7.12} = \frac{-15}{7.12} \approx -2.11. \]

Step 4: Find the Probability from Z-Table

Using the Z-table, find the probability for \( Z = -2.11 \). The Z-table value for \( Z = -2.11 \) corresponds to approximately \( 0.0174 \) or 1.74\%.

Step 5: Interpret the Probability

The probability obtained, \( 0.0174 \), represents the likelihood of observing a sample mean of \( 335 \) or less. Since we need the likelihood of finding a sample mean of \( 335 \) or larger, we calculate \( 1 - 0.0174 = 0.9826 \).

Key Concepts

Standard Error

The **Standard Error (SE)** is a crucial concept in statistics, especially when dealing with a sample from a population. It helps us estimate the precision of the sample mean in relation to the population mean. The formula for the standard error is:\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]where:

• \( \sigma \) is the population standard deviation
• \( n \) is the sample size

In our exercise, the population's standard deviation is \( \\(45 \), and the sample size is 40 women. By substituting these into the formula, we can find the standard error:\[ \text{SE} = \frac{45}{\sqrt{40}} \approx 7.12 \]This result tells us that the sample mean is likely to vary by about \( \\)7.12 \) from the population mean just due to random sampling variation. Understanding the standard error gives us a sense of how much the sample mean is likely to differ from the actual population mean. It's fundamental for constructing confidence intervals and conducting hypothesis tests.

Z-Score

The **Z-Score** is a measure that describes a value's relationship to the mean of a group of values. When you calculate the Z-score, you standardize it, meaning you measure how many standard deviations an element is from the mean. The Z-score formula for our sample mean is:\[ Z = \frac{\bar{x} - \mu}{\text{SE}} \]where:

• \( \bar{x} \) is the sample mean
• \( \mu \) is the population mean
• \( \text{SE} \) is the standard error

In the exercise, substituting the values gives us:\[ Z = \frac{335 - 350}{7.12} \approx -2.11 \]This result indicates that the sample mean of \( \\(335 \) is more than two standard deviations below the population mean of \( \\)350 \). The Z-score allows us to use the standard normal distribution to examine the likelihood of our observed sample mean.

Probability Calculation

In statistics, **Probability Calculation** is used to find the likelihood of an event occurring under the normal distribution curve. Once we have the Z-score, we can determine how likely it is to get a sample mean as extreme as ours. We use a Z-table, which gives the probability that a statistic is less than or equal to a Z-value. For our calculated Z-score of \(-2.11\), the table gives a probability of approximately 0.0174 or 1.74%. However, our original question asks for the probability of finding a mean of \(335\) or larger. Since the Z-table gives us the probability for \(335\) or smaller, we must calculate:\[ 1 - 0.0174 = 0.9826 \]which means there is a 98.26% chance of observing a sample mean of \(335\) or larger. This high probability suggests that a sample mean less than the population mean is relatively uncommon under the given normal distribution model.

Sample Mean Analysis

**Sample Mean Analysis** is the process of using a sample to make inferences about a population mean. By taking a sample of 40 women out of the entire population, we obtain a sample mean which is then used to understand more about the spending behavior of the typical 50-year-old woman. Here's how it works:- **Population Mean:** This is a known value, provided as \(\\(350\).- **Sample Mean:** Calculated from the sample data, here it is \(\\)335\).- **Analysis:** Through our calculations, we identified that the sample mean is lower than the population mean, but there's still a high probability (98.26%) of such a result occurring by chance.This analysis helps us understand whether our sample reflects the population well or if there might be a significant deviation. With a larger sample, our sample mean would be more accurate. However, even with 40 women, our findings reveal informative insights about the population's spending habits on personal care products.