Questions: Use the following sample to estimate a population mean μ. 39.8 42 58.4 23.5 40.5 47.8 48.2 13.1 48.1 22.6 69.2 50.5 51.6 78.5 28.2 12.2 69.1 38.9 13.8 63.5 21.6 25.8 32.4 54.4 15.9 17 47.2 36.6 57.8 64.8 30.2 27.5 56.7 25.4 46.4 58.3 36.7 38.9 69.1 27.6 30.7 50.1 30 55.5 42.3 27 47.2 50.2 48.4 53.2 63.9 44 26.9 70 45.8 23.6 41.6 42.9 37 55 43.8 Find the 95% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. < μ <

Solution Steps

To find the 95% confidence interval for the population mean, we need to follow these steps:

1. Calculate the sample mean (\(\bar{x}\)).
2. Calculate the sample standard deviation (s).
3. Determine the sample size (n).
4. Use the t-distribution to find the critical value for a 95% confidence level.
5. Calculate the margin of error using the formula: \( \text{Margin of Error} = t \times \frac{s}{\sqrt{n}} \).
6. Determine the confidence interval using the formula: \( \bar{x} \pm \text{Margin of Error} \).

The sample mean (\(\bar{x}\)) is calculated as: \[ \bar{x} = 42.2770 \]

The sample standard deviation (\(s\)) is calculated as: \[ s = 16.0547 \]

The sample size (\(n\)) is: \[ n = 61 \]

For a 95% confidence level and \(n-1\) degrees of freedom, the t-critical value (\(t\)) is: \[ t = 2.0003 \]

The margin of error (ME) is calculated using the formula: \[ \text{ME} = t \times \frac{s}{\sqrt{n}} = 2.0003 \times \frac{16.0547}{\sqrt{61}} = 4.1118 \]

The 95% confidence interval for the population mean (\(\mu\)) is: \[ \bar{x} \pm \text{ME} = 42.2770 \pm 4.1118 \] Thus, the confidence interval is: \[ (38.1653, 46.3888) \]

Final Answer