Question

A random sample of 121 students from the University of Oklahoma had a sample mean ACT score of 23.4 with a sample standard deviation of 3.65. Construct a 95% confidence interval for the population mean ACT score of University of Oklahoma students.

Answer 1

Answer:

(22.74,24.06) is the required 95% confidence interval for the population mean ACT score of University of Oklahoma students.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 121

Sample mean = 23.4

Sample standard deviation = 3.65

Level of significance = 0.05

Degree of freedom

$= n - 1 = 120$

95% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Calculation of critical value:

$t_{critical}\text{ at degree of freedom 120 and}~\alpha_{0.05} = \pm 1.9799$  

Putting the values, we get,  

$23.4\pm 1.9799(\dfrac{3.65}{\sqrt{121}} )\\\\ = 23.4 \pm 0.6569\\\\ = (22.7431 ,24.0569)\approx (22.74,24.06)$  

(22.74,24.06) is the required 95% confidence interval for the population mean ACT score of University of Oklahoma students.