Answer:
Option C) 5.62 points
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 20
Sample mean score = 64
Sample standard deviation, s = 12
Degree of freedom =
[tex]=n-1\\=20-1\\=19[/tex]
We have to calculate margin of error for a 95% confidence interval.
Formula for margin of error:
[tex]t_{critical}\times \displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.05} = \pm 2.093[/tex]
[tex]2.093\times (\displaystyle\frac{12}{\sqrt{20}} )\\\\ =5.6161\approx 5.62[/tex]
Thus, the correct answer is
Option C) 5.62 points
