Problem 20
Question
Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9. Develop a 99 percent confidence interval for the population mean.
Step-by-Step Solution
Verified Answer
The 99% confidence interval is (36.54, 43.46).
1Step 1: Determine the Sample Size (n) and Degrees of Freedom
In this problem, the sample size \( n = 49 \). The degrees of freedom for the t-distribution when estimating the mean is \( n - 1 = 48 \).
2Step 2: Identify Sample Mean (\( \bar{x} \)) and Standard Deviation (s)
The problem provides that the sample mean \( \bar{x} = 40 \) and the sample standard deviation \( s = 9 \).
3Step 3: Select Confidence Level and Corresponding Critical Value
A 99% confidence level is required. For a t-distribution with 48 degrees of freedom, consult a t-table or use statistical software to find the critical value, \( t^* \), which is approximately 2.684.
4Step 4: Calculate the Standard Error of the Mean (SEM)
The standard error of the mean (SEM) is given by the formula \( \text{SEM} = \frac{s}{\sqrt{n}} \). Substituting the given values: \( \text{SEM} = \frac{9}{\sqrt{49}} = \frac{9}{7} = 1.29 \).
5Step 5: Calculate the Confidence Interval
The confidence interval is calculated using the formula \( \bar{x} \pm t^* \times \text{SEM} \). Plugging in the values: \[ 40 \pm 2.684 \times 1.29 \]. This computes to \[ 40 \pm 3.46 \]. Therefore, the confidence interval is \( (36.54, 43.46) \).
Key Concepts
Sample SizeDegrees of FreedomStandard Error of the MeanCritical Value
Sample Size
When conducting statistical analyses, the term "sample size" refers to the number of observations or data points collected in a study. In the context of developing a confidence interval, the sample size significantly influences the precision and reliability of the interval estimate. A larger sample size generally leads to more precise estimates since it reduces the standard error.
- In our example, the sample size is \( n = 49 \).
- The sample was drawn from a larger population of 500 items, making \( n = 49 \) adequately representative assuming random sampling.
- Remember, the sample size is crucial in determining the degrees of freedom, which is essential for selecting the right statistical method to use, like the t-distribution used in this case.
Degrees of Freedom
Degrees of freedom is a statistical concept that describes the number of values in a calculation that are free to vary. For a sample mean, the degrees of freedom is typically given by \( n - 1 \), where \( n \) is the sample size.
- With a sample size of 49, we find the degrees of freedom by calculating \( n - 1 = 48 \).
- Degrees of freedom is critical when using the t-distribution to determine the critical value necessary for constructing a confidence interval.
Standard Error of the Mean
The standard error of the mean (SEM) represents the variability of the sample mean estimate from the true population mean. It assesses how far the sample mean is expected to deviate from the population mean and is calculated using the formula \( \text{SEM} = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation.
- In the given problem, the standard deviation is 9 and the sample size is 49, leading to an SEM of \( \frac{9}{7} = 1.29 \).
- The SEM is crucial for determining the width of the confidence interval: a smaller SEM results in a narrower interval, indicating a more precise estimate.
Critical Value
The critical value corresponds to the point on the t-distribution that defines the margins of the confidence interval for a specified confidence level. This critical value, often denoted as \( t^* \), is derived based on the desired level of confidence and the degrees of freedom.
- For a 99% confidence level and 48 degrees of freedom, the critical value from the t-distribution table is approximately 2.684.
- The critical value helps to scale the standard error, thus determining the distance from the sample mean to the interval boundaries.
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