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?
oa. 28.2<μ<30.8
ob. 27.9<μ<31.1
oc. 28.4<μ<30.6
od. 27.8<μ<31.2

Explanation:

Step1: Find the critical value

For a 90% confidence interval, the significance level \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). Using the standard normal distribution \(z_{\alpha/2}=z_{0.05}\approx1.645\) (from standard normal - distribution tables).

Step2: Calculate the margin of error

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

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Step3: Calculate the confidence interval

The sample mean \(\bar{x}=29.5\). The confidence interval is \(\bar{x}-E<\mu <\bar{x} + E\).

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Answer:

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