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Magazine Chapter 9: Problem 54
Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the mean annual health insurance premium for a family with coverage through an employer was \(\$ 10,979 .\) The standard deviation of the sample was \(\$ 1,000 .\) a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within \(\$ 250\) at 99 percent confidence?
Short Answer
Expert verified
a. 90% CI is (\$10,592.22, \$11,365.78). b. Sample of 107 needed.
Step by step solution
01
Identify Given Information Part A
For part (a), we start by identifying key information:- Sample Mean (\( \bar{x} \)) = \\(10,979- Standard Deviation (\( s \)) = \\)1,000- Sample Size (\( n \)) = 20- Confidence Level = 90%Since the sample size is small, we'll use the t-distribution to compute the confidence interval.
02
Find t-Value for 90% Confidence Interval
For a 90% confidence level and a sample size of 20, the degrees of freedom is \( n-1 = 19 \). Using a t-table, find the critical t-value for 19 degrees of freedom at 90% confidence, which is approximately 1.729.
03
Calculate Margin of Error Part A
The margin of error (E) is calculated using the formula:\[E = t \times \frac{s}{\sqrt{n}}\]Substitute the values:\[E = 1.729 \times \frac{1,000}{\sqrt{20}} \approx 386.78\]
04
Compute the 90% Confidence Interval Part A
The confidence interval is given by:\[(\bar{x} - E, \bar{x} + E)\]Substituting the values:\[(10,979 - 386.78, 10,979 + 386.78) = (10,592.22, 11,365.78)\]Thus, the 90% confidence interval is approximately \(\\(10,592.22\) to \(\\)11,365.78\).
05
Identify Given Information Part B
For part (b), we need to determine the sample size needed to find the population mean within \(\\(250\) at a 99% confidence level. The key pieces of information include:- Desired Margin of Error (E) = \\)250- Confidence Level = 99%- Standard Deviation (\( s \)) = \$1,000.
06
Find z-Value for 99% Confidence Interval
Since the sample size needs to be determined, we use the z-distribution, and for a 99% confidence level, the z-value is approximately 2.576.
07
Calculate Required Sample Size Part B
The sample size (n) is determined using the formula:\[n = \left( \frac{z \cdot s}{E} \right)^2\]Substitute the known values:\[n = \left( \frac{2.576 \times 1,000}{250} \right)^2 = 106.09\]Since the sample size must be a whole number, round up to 107.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sample Size Determination
Determining the required sample size is a crucial aspect of statistical analysis. When you want to estimate a population mean with a certain level of confidence and a specific margin of error, you need to calculate the minimum number of observations required. This helps ensure your estimates are reliable and accurate. Let's break it down:
- Desired Margin of Error (E): This is the maximum allowable difference between the sample mean and the true population mean. It acts as a cushion for your estimate.
- Confidence Level: This reflects how sure you want to be about your results. Higher confidence levels require larger sample sizes.
- Standard Deviation (s): A measure of the data's spread. More variability means you need more data to capture the true mean accurately.
- Finally, use the formula for calculating sample size: \[n = \left( \frac{z \cdot s}{E} \right)^2\]To find the required sample size, adjust the desired confidence level and margin of error to meet your specific requirements. For part b in the exercise, the calculation shows a sample size of 107 to find the mean within $250 at 99% confidence.
t-Distribution
In scenarios where the sample size is small (typically less than 30), or the population standard deviation is unknown, the t-distribution is used instead of the z-distribution. It accounts for the increased variability expected with smaller samples. Here are key points about the t-distribution:
- Degrees of Freedom: This is calculated as the sample size minus one (i.e., \(n-1\)). This adjusts the distribution's shape, making it more spread out with smaller samples.
- Critical t-Value: Obtained from the t-table, it shows how extreme the sample result must be to be considered unusual. It's used in calculating the margin of error.
- The t-distribution is more conservative, leading to wider confidence intervals compared to the z-distribution, reflecting the greater uncertainty with small samples.
- In the given exercise, with a sample size of 20, the degrees of freedom are 19, and a t-value of approximately 1.729 is used for the 90% confidence interval.
Margin of Error
The margin of error represents the range within which the true population parameter is expected to lie. This range provides a buffer around the sample estimate, acknowledging possible sampling variability. Key components include:
- Calculation: The margin of error is calculated as the critical value (t or z) times the standard deviation divided by the square root of the sample size: \[E = \text{value} \times \frac{s}{\sqrt{n}}\]
- Influence Factors: It is affected by the confidence level (higher levels increase margin of error), sample size (larger samples reduce margin), and data variability (greater variability increases margin).
- In the exercise, for a 90% confidence interval, the margin of error is found to be approximately 386.78, reflecting possible differences from the actual population mean.
Z-Distribution
The z-distribution is typically used when you have a large sample size or know the population standard deviation. It assumes a normal distribution of the sample data, allowing for more precise confidence interval estimation under these conditions. Understanding the z-distribution involves:
- Use with Larger Samples: When sample sizes are large (\(n \geq 30\)), the Central Limit Theorem suggests that the sampling distribution of the sample mean is approximately normal.
- Critical z-Value: Like the t-value, the z-value represents the number of standard deviations a point is from the mean. For confidence intervals, typical critical values include 1.96 for 95%, 2.576 for 99%, etc.
- In part b of the exercise, the z-distribution is used to find the sample size for estimating the mean within $250 at 99% confidence, with a critical value of approximately 2.576.
