Problem 35 - Basic Statistics for Business and Economics with Student

Question

Marty Rowatti recently assumed the position of director of the YMCA of South Jersey. He would like some current data on how long current members of the YMCA have been members. To investigate, suppose he selects a random sample of 40 current members. The mean length of membership of those included in the sample is 8.32 years and the standard deviation is 3.07 years. a. What is the mean of the population? b. Develop a 90 percent confidence interval for the population mean. c. The previous director, in the summary report she prepared as she retired, indicated the mean length of membership was now "almost 10 years." Does the sample information substantiate this claim? Cite evidence.

Step-by-Step Solution

Answer

b. The 90% confidence interval is (7.503, 9.137). c. The sample does not support the claim of a 10-year membership mean.

Step 1: Determine the sample mean

The problem states that the mean length of membership for the sample is \( \overline{x} = 8.32 \) years. This is the mean of the sample, not the population, as the population mean is unknown.

Step 2: Identify the known information for confidence interval

We are given the sample standard deviation \( s = 3.07 \) years, and the sample size \( n = 40 \). We will use these to develop the confidence interval for the population mean, since the population standard deviation is unknown.

Step 3: Calculate the standard error of the mean

The standard error (SE) of the mean is calculated as \( SE = \frac{s}{\sqrt{n}} \). Substituting the known values:\[ SE = \frac{3.07}{\sqrt{40}} \approx 0.485 \]

Step 4: Find the appropriate t-score for a 90% confidence level

For a 90% confidence level and 39 degrees of freedom \((n-1 = 40-1)\), we refer to a t-distribution table to find that the critical t-value \( t^* \approx 1.685 \).

Step 5: Calculate the 90% confidence interval for the population mean

The confidence interval is given by \( \overline{x} \pm t^* \times SE \). Substitute the known values:\[ 8.32 \pm 1.685 \times 0.485 \]This gives the interval \( 8.32 \pm 0.817 \), which is \( (7.503, 9.137) \).

Step 6: Compare the confidence interval to the previous director's claim

The interval \( (7.503, 9.137) \) does not include 10, as the previous director claimed the mean was "almost 10 years". This suggests that the sample data do not support the claim that the mean length of membership is close to 10 years.

Key Concepts

Mean of Population

The mean of the population, often denoted as \( \mu \), represents the average of a particular characteristic across the entire population. In this exercise, the population refers to all current members of the YMCA of South Jersey and the characteristic of interest is the length of membership.

Calculating the mean of the population is often challenging because gathering data from every single member (or unit) within the population is typically impractical due to time and resource constraints.

Instead, we draw a sample - a smaller, manageable group - and use its data to make estimates about the population mean. The sample provides a sample mean \( \overline{x} \), which serves as an unbiased estimate of the unknown population mean. In summary:

• The population mean is the true average we aim to estimate.
• In our example, the population mean is unknown, but we have a sample mean of 8.32 years.

Sample Mean

The sample mean, denoted as \( \overline{x} \), is the average of a set of observations collected from a sample of the population. It provides an estimate of the population mean, especially when the population mean \( \mu \) is unknown or difficult to determine.

In this scenario, the director collected data from 40 random members, yielding a sample mean of 8.32 years. This number tells us, on average, how long members in this sample have been with the YMCA.

Key points about the sample mean include:

• It offers a "snapshot" of the population mean.
• While it is an estimate, it may differ due to random sample variations.
• Used in further calculations, such as developing confidence intervals to estimate the population mean.

Standard Deviation

Standard deviation is a measure of variation or spread in a set of data. For a sample, this is shown by \( s \), which tells us how much individual data points (like years of membership) deviate from the sample mean.

In Marty's case, the sample standard deviation is 3.07 years. This suggests that the individual membership durations vary, on average, by 3.07 years from the sample mean of 8.32 years.

Understanding standard deviation is important because:

• It provides insights into the consistency within the sample data.
• A larger standard deviation implies more spread, whereas a smaller one indicates that data points are closer to the mean.
• It forms a critical component of the formula for calculating the confidence interval.

t-Distribution

When determining confidence intervals for the mean with unknown population standard deviation and smaller samples, we rely on the t-distribution. This is a probability distribution resembling the normal distribution but accounts for extra variability due to small sample sizes.

For this exercise, we have:

• Sample size of 40, leading to 39 degrees of freedom \((n-1)\).
• The t-distribution is used to determine the t-score, which adjusts our confidence interval range given the sample size.
• For our 90% confidence level, a t-score of approximately 1.685 is applied.

Using the t-distribution helps us determine:

• How precise our estimate of the population mean is likely to be.
• A credible range (confidence interval) within which the actual population mean lies with a certain level of confidence.