Questions: Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean. 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.3 pounds and a standard deviation of 2.4 pounds. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service? (A) 9.4 lb<μ<11.2 lb (B) 9.6 lb<μ<11.0 lb (C) 9.3 lb<μ<11.3 lb (D) 9.5 lb<μ<11.1 lb

Solution Steps

To find the confidence interval for the population mean, we use the formula for the confidence interval of the mean with a known sample standard deviation. Since the sample size is 37, which is greater than 30, we can use the Z-distribution. The formula is:

\[ \text{CI} = \bar{x} \pm Z \left(\frac{s}{\sqrt{n}}\right) \]

where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(Z\) is the Z-score corresponding to the desired confidence level (for 95%, \(Z \approx 1.96\)).

We are given the following values:

• Sample mean \(\bar{x} = 10.3\)
• Sample standard deviation \(s = 2.4\)
• Sample size \(n = 37\)
• Confidence level = 95%

For a 95% confidence level, the Z-score is approximately \(Z = 1.960\).

The margin of error (ME) is calculated using the formula: \[ \text{ME} = Z \left(\frac{s}{\sqrt{n}}\right) \] Substituting the given values: \[ \text{ME} = 1.960 \left(\frac{2.4}{\sqrt{37}}\right) \approx 0.7733 \]

The confidence interval (CI) is given by: \[ \text{CI} = \bar{x} \pm \text{ME} \] Substituting the values: \[ \text{Lower bound} = 10.3 - 0.7733 \approx 9.5267 \] \[ \text{Upper bound} = 10.3 + 0.7733 \approx 11.0733 \]

Final Answer