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, $\mu$ , of all students taking the test?
A) $28.2

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, $\mu$ , of all students taking the test?
A) $28.2 < \mu < 30.8$
B) $28.4 < \mu < 30.6$
C) $27.8 < \mu < 31.2$
D) $27.9 < \mu < 31.1$

Quizplus · Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (z * (standard deviation/√n)). For 90% confidence, z is approximately 1.645. Plugging in the values: 29.5 ± (1.645 * (5.2/√59)) gives the interval 28.4 to 30.6.

A Group of 59 Randomly Selected Students Have a Mean μ\muμ

A Group of 59 Randomly Selected Students Have a Mean $\mu$