Question

the monthly incomes for 12 randomly selected people, each with a bachelors degree in economics, are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 4450.97 4596.67 4366.39 4455.64 4151.92 3727.05 4283.53 4527.91 4407.37 3946.65 4023.24 4221.57 (a) find the sample mean. \\( \overline { x } = 4263.2 \\) (round to one decimal place as needed.) (b) find the sample standard deviation. \\( s = 260.2 \\) (round to one decimal place as needed.) (c) construct a 99% confidence interval for the population mean \\( \mu \\). a 99% confidence interval for the population mean is (). (round to one decimal place as needed.)

Explanation:

Step1: Determine the critical value

Since the population standard deviation is unknown and the sample size \(n = 12\), we use the \(t\) - distribution. The degrees of freedom \(df=n - 1=12-1 = 11\). For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\). Using a \(t\) - table or calculator, \(t_{\frac{\alpha}{2},df}=t_{0.005,11}=3.106\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\). We know that \(s = 260.2\), \(n = 12\), and \(t_{\frac{\alpha}{2}}=3.106\).

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Step3: Construct the confidence interval

The formula for the confidence interval for the population mean \(\mu\) when \(\sigma\) is unknown is \(\bar{x}-E<\mu<\bar{x} + E\). We know that \(\bar{x}=4263.2\) and \(E = 233.4\)

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Answer:

\((4029.8,4496.6)\)