Problem 50 - Basic Statistics for Business & Economics

Question

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged 10,979 . The standard deviation of the sample was 1,000 . a. Based on this sample information, develop a \(90 \%\) confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within 250$ at 99% confidence?

Step-by-Step Solution

Answer

a) CI: [10,593, 11,365]. b) Sample size: 107.

Step 1: Identify Given Information

We have a sample mean \( \bar{x} = 10,979 \), a sample standard deviation \( s = 1,000 \), and the sample size \( n = 20 \). We need to find a 90% confidence interval for the population mean.

Step 2: Find Critical Value for 90% Confidence Interval

Since the sample size is small, we use the t-distribution. The degrees of freedom \( df = n - 1 = 19 \). Using a t-table or calculator, we find the critical value \( t_{0.05, 19} \approx 1.729 \) for a 90% confidence level.

Step 3: Calculate Margin of Error

The margin of error (E) is calculated by \( E = t \frac{s}{\sqrt{n}} \). Substitute the values: \( E = 1.729 \frac{1,000}{\sqrt{20}} \approx 386 \).

Step 4: Determine the Confidence Interval

The confidence interval is given by \( \bar{x} \pm E \). Thus, it is \( 10,979 \pm 386 \). This results in the interval \( [10,593 , 11,365] \).

Step 5: Find the Required Sample Size for Part b

Use the sample size formula for the population mean: \( n = \left( \frac{Z \cdot s}{E} \right)^2 \). At 99% confidence, the Z-value is approximately 2.576 and \( E = 250 \). Substitute the values: \( n = \left( \frac{2.576 \cdot 1,000}{250} \right)^2 = 106.36 \).

Step 6: Round Up Sample Size

Since the sample size must be a whole number, round up \( 106.36 \) to get \( n = 107 \).

Key Concepts

Population Mean

When discussing a confidence interval, the term "population mean" refers to the average value of a certain characteristic for an entire population. In our example, this characteristic is the yearly premium for health insurance policies. The aim is to estimate the average premium if we could survey every family that has insurance through their employer. However, since surveying the entire population is usually impractical, we use a smaller sample to estimate this mean.

By surveying 20 families, we calculated an average (or mean) premium of $10,979. This sample mean (\[\bar{x}\ = 10,979\]) serves as a point estimate for the population mean. We want to determine how precise this estimate is, which is why we use confidence intervals.

Sample Size Determination

When determining the sample size needed to estimate the population mean with a certain precision, we consider several factors:

• Desired Margin of Error (E): This represents how close we want our sample mean to be to the true population mean. In our scenario, we want our estimate to lie within $250 of the actual mean.
• Confidence level: A higher confidence level requires a larger sample size. For a 99% confidence level, as in our example, we use a Z-value of approximately 2.576.
• Sample standard deviation (s): This measures the spread of our data. A larger standard deviation means more variation, which often calls for a larger sample size.

By plugging these values into the sample size formula: \( n = \left( \frac{Z \cdot s}{E} \right)^2 \), we determine that a sample size of 107 is needed. This accounts for the fact that calculations should result in whole numbers since we cannot survey a fraction of a person.

t-Distribution

The t-distribution is a key ingredient when estimating a population mean, especially when dealing with small sample sizes (typically less than 30). Unlike the normal distribution, the t-distribution takes into account the added uncertainty that comes with smaller samples, which often leads to wider confidence intervals. It is slightly flatter and more spread out than the normal curve.

For our exercise, since the sample size is 20, we use the t-distribution. We calculate the degrees of freedom (df) as \( n - 1 \), which in this case is 19. The critical value used in the confidence interval formula can be found using a t-table or calculator, and for a 90% confidence interval, it is \( t_{0.05, 19} \approx 1.729\). This value helps adjust for the sample size, providing a more accurate estimate of the population mean's confidence interval.

Margin of Error

The margin of error quantifies the uncertainty in our estimate of the population mean. It defines the range within which we expect the true population mean to fall, given a certain level of confidence. In mathematical terms, it is the product of the critical value and the standard error of the mean.

Our margin of error (E) is calculated using the formula:\[ E = t \cdot \frac{s}{\sqrt{n}} \]With our values plugged in, it became:\[ E = 1.729 \cdot \frac{1,000}{\sqrt{20}} \approx 386 \]This means that we consider our sample mean can vary by ±386 when extending this to estimate the population mean at a 90% confidence level.

Thus, the confidence interval is the sample mean plus or minus the margin of error, resulting in the interval [10,593, 11,365]. This solidifies that we expect the true average yearly premium to lie in this range, offering clarity on how precise our estimate is.