Question

a group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. what is the 90% confidence interval for the mean score, μ, of all students taking the test?
a. 28.2 < μ < 30.8
b. 27.8 < μ < 31.2
c. 28.4 < μ < 30.6
d. 27.9 < μ < 31.1

Explanation:

Step1: Find the critical value

For a 90% confidence interval with \(n = 59\) (so \(n-1=58\) degrees of freedom, and since \(n = 59>30\), we can approximate using the standard normal distribution). The critical value \(z_{\alpha/2}\) for a 90% confidence interval (\(\alpha=1 - 0.90=0.10\), \(\alpha/2 = 0.05\)) is \(z_{0.05}\approx1.645\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 5.2\), \(n = 59\)

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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 \(\bar{x}=29.5\)

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Answer:

C. \(28.4<\mu<30.6\)