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. 27.9<μ<31.1
d. 28.4<μ<30.6

Explanation:

Step1: Determine the critical value

For a 90% confidence interval, the significance level \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2 = 0.05\).
Since \(n = 59\) (\(n>30\)), we use the standard normal distribution.
Looking up in the standard - normal table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\)
Given \(s = 5.2\), \(n = 59\)
\(E=1.645\times\frac{5.2}{\sqrt{59}}\)
\(\sqrt{59}\approx7.681\), \(\frac{5.2}{7.681}\approx0.677\)
\(E = 1.645\times0.677\approx1.114\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=29.5\)
\(\bar{x}-E=29.5 - 1.114=28.386\approx28.4\)
\(\bar{x}+E=29.5 + 1.114=30.614\approx30.6\)

Answer:

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