Problem 38 - Basic Statistics for Business and Economics with Student

Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is \(\$ 23.50\) with a standard deviation of \(\$ 5.00 .\) Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. a. What is the likelihood the sample mean is at least \(\$ 25.00 ?\) b. What is the likelihood the sample mean is greater than \(\$ 22.50\) but less than \(\$ 25.00 ?\) c. Within what limits will 90 percent of the sample means occur?

Step-by-Step Solution

Answer

a) About 1.7%, b) About 90.4%, c) \([22.34, 24.66]\).

Step 1: Understanding the Central Limit Theorem

Since the sample size is 50, which is a reasonable size, the Central Limit Theorem (CLT) can be applied. According to the CLT, the sampling distribution of the sample mean will be approximately normally distributed with the mean equal to the population mean, \( \mu = \\( 23.50 \), and the standard deviation (standard error) equal to \( \sigma / \sqrt{n} \), where \( \sigma = \\) 5.00 \) and \( n = 50 \).

Step 2: Calculating the Standard Error

The standard error of the sample mean is calculated as: \( \text{SE} = \sigma / \sqrt{n} = 5.00 / \sqrt{50} \approx 0.7071 \).

Step 3: Solving Question a - Likelihood Sample Mean \( \geq \$ 25.00 \)

Convert \( \$ 25.00 \) to a Z-score using the formula: \( Z = (X - \mu) / \text{SE} \). Thus, \( Z = (25.00 - 23.50) / 0.7071 \approx 2.12 \). Using the Z-table or standard normal distribution, find \( P(Z \geq 2.12) \), which is approximately \( 0.017 \).

Step 4: Solving Question b - Likelihood Sample Mean \( 22.50 \lt X \lt 25.00 \)

First, calculate the Z-score for \( \$ 22.50 \): \( Z = (22.50 - 23.50) / 0.7071 \approx -1.41 \). The probability for this range is \( P(-1.41 < Z < 2.12) \), calculated by finding \( P(Z < 2.12) \approx 0.983 \) and \( P(Z < -1.41) \approx 0.079 \). Subtract these probabilities: \( 0.983 - 0.079 = 0.904 \).

Step 5: Solving Question c - 90% Confidence Interval

To find the limits within which 90% of sample means fall, use the formula for the confidence interval: \( \mu \pm Z* \cdot \text{SE} \), where \( Z* \approx 1.645 \) for 90% confidence. The interval is \( 23.50 \pm 1.645 \times 0.7071 \), which results in \([22.34, 24.66]\).

Step 6: Compile the Answers

The likelihood that the sample mean is at least \( \\( 25.00 \) is approximately 1.7%. The likelihood that the sample mean is between \( \\) 22.50 \) and \( \\( 25.00 \) is approximately 90.4%. The limits within which 90% of the sample means occur are approximately \( \\) 22.34 \) to \( \$ 24.66 \).

Key Concepts

Sampling Distribution

When we talk about sampling distributions, we're considering how the sample means from various samples of the same size will be distributed. Think of it as a way to summarize how different sample averages might look if we repeatedly took samples from a population.
In our context, the sampling distribution of the mean involves samples of customer purchases at Churchill's Grocery Store. Since we know the population mean is $23.50 and the standard deviation is $5.00, we can use these to analyze the sample mean's distribution. The Central Limit Theorem plays a crucial role here, telling us that the sampling distribution of the sample mean will be approximately normal if the sample size is large enough. Here, a sample size of 50 is sufficient for the theorem to apply.
Now, this sampling distribution will have the same mean as the population, $23.50, but the standard deviation will be the standard error, which is smaller because it accounts for the sample size. This helps us calculate the likelihood of different sample mean outcomes, such as whether it's greater or less than certain values.

Confidence Interval

A confidence interval provides a range of values within which we expect a population parameter, like the mean purchase amount, to fall. In practice, this means we're estimating a range where the true mean of our population might lie, based on our sample data.
For instance, in our problem, we calculated a 90% confidence interval for the sample means. This range, $22.34 to $24.66, is where we expect 90% of sample mean purchase amounts to fall. It uses the normal distribution properties along with the calculated standard error to determine how far we can expect sample means to deviate from the population mean.

• The Z-score plays a crucial role here, translating a desired confidence level (90% in this case) into a number that helps in scaling the standard error.
• The wider the interval, the more confident we are about where the population mean might land, but we also get less precise the wider we go.

Confidence intervals are helpful because they provide a way to quantify the uncertainty in our sample estimates.

Standard Error

The standard error (SE) is a measure of how spread out the sampling distribution is, specifically for the sample means. It's derived from the population standard deviation, adjusted by the square root of the sample size. It essentially gives us an indication of the "average" amount by which the sample means will differ from the actual population mean.
In this context, the standard error is calculated as SE = 5.00 / \(\sqrt{50}\) which is approximately 0.7071. This value tells us how much variability we can expect among the sample means compared to the actual population mean. A smaller standard error implies that the sample means are closely clumping around the population mean.
Recognizing the importance of standard error helps us understand confidence intervals and the results of hypothesis tests. It is key to realizing how much "noise" versus "signal" there is in our data, leading to more informed statistical decisions. This is why, in statistical analyses, reducing the standard error (often by increasing sample size) can lead to sharper, more reliable insights.