Answer to Question #327628 in Statistics and Probability for Loloy

Question

The following data represent a random sample of 15 marks out of 20) on a Statistics quiz. Assume that the marks are normally distributed 4 7 7 5 3 10 6 8 7 9 8 9 5 4 7 a. Determine the standard deviation of the marks. b. Estimate the population mean with 95% confidence.

Answer 1

First let's find mean:

$\mu=\frac{1}{15}(4+7+7+5+3+10+6+8+\\ +7+9+8+9+5+4+7)=\\ =\frac{1}{15}(3+2\cdot4+2\cdot5+6+\\ 4\cdot7+2\cdot8+2\cdot9+10)=6.6$

a) Variance:

$\sigma^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\mu)^2=\\ =\frac{1}{14}(3.6^2+2\cdot2.6^2+2\cdot1.6^2+0.6^2+\\ +4\cdot0.4^2+2\cdot1.4^2+2\cdot2.4^2+3.4^2)\approx4.257$

Standard deviation:

$\sigma=\sqrt{\sigma^2}\approx2.063$

b) t-value for 15 - 1 = 14 degrees of freedom and 95% confidence is 2.145

Confidence interval:

$\mu\pm t\frac{\sigma}{\sqrt{n}}=6.6\pm 2.145\frac{2.063}{\sqrt{15}}=6.60\pm 1.14$