Question

construct a 99% confidence interval for the population mean, μ. assume the population has a normal distribution. a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.
a. (19.9,24.9)
b. (18.7,24.1)
c. (16.3,26.9)
d. (17.2,23.6)

Explanation:

Step1: Determine the critical value

Since the population standard deviation is unknown and \(n = 19\) (small - sample, \(n<30\)), we use the \(t\) - distribution. The degree of freedom \(df=n - 1=19 - 1 = 18\). For a \(99\%\) confidence interval, \(\alpha=1 - 0.99 = 0.01\), and \(\frac{\alpha}{2}=0.005\). Looking up in the \(t\) - table, \(t_{\frac{\alpha}{2},df}=t_{0.005,18}=2.878\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 3.8\), \(n = 19\).

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=22.4\)

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Answer:

A. \((19.9,24.9)\)