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:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 64 and}~\alpha_{0.10} = \pm 1.669[/tex]
[tex]21.4 \pm 1.669(\frac{3.2}{\sqrt{65}} ) = 21.4 \pm 0.6624 = (20.7376 ,22.0624)[/tex]
