Question

a random sample of 65 high school students has a norma distribution. The smple mean average ACT exam score was 21.4 with a 3.2 sample standard deviation. Construct a 90% confidence interval estimate of the population mean average ACT exam and write a sentence to explain the confidenve interval.

Answer 1

Answer:

90% Confidence interval:   (20.7376 ,22.0624)

Explanation:

We are given the following information in the question:

Mean, μ = 21.4

Standard Deviation, σ = 3.2

Sample size, n = 65

We are given that the distribution of ACT exam score is a bell shaped distribution that is a normal distribution.

90% Confidence interval:  

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

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 64 and}~\alpha_{0.10} = \pm 1.669$ 

$21.4 \pm 1.669(\frac{3.2}{\sqrt{65}} ) = 21.4 \pm 0.6624 = (20.7376 ,22.0624)$