Question

construct a 95% confidence interval for the population mean, μ. assume the population has a normal distribution. a sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2.

a. (87.12,98.32)
b. (66.35,69.89)
c. (56.12,78.34)
d. (77.29,85.71)

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 25\) (small - sample, \(n<30\)) and the population is normally distributed, we use the \(t\) - distribution. The degree of freedom \(df=n - 1=25-1 = 24\). For a \(95\%\) confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\). Looking up in the \(t\) - table, \(t_{\frac{\alpha}{2},df}=t_{0.025,24}=2.064\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 10.2\) (sample standard deviation) and \(n = 25\).

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=81.5\) (sample mean).

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Answer:

D. \((77.29,85.71)\)