Question

A normal population with unknown variance is believed to have a mean of 20. Is on likely a random sample of size 9 from this population that has mean 24 and standard deviation of 4.1. If not, what conclusion would you draw

Answer 1

$H_0: \mu=20 \\ H_1: \mu ≠20 \\ n= 9 \\ \bar{x} = 24 \\ s = 4.1$

Degree of freedom df = n-1 = 8

Test-statistic:

$t = \frac{\bar{x} - \mu }{s/ \sqrt{n}} \\ t = \frac{24-20}{4.1 / \sqrt{9}} = 2.93 \\ P-value = P(|t| > 2.93) = P(t< -2.93) + P(t> 2.93) = 0.019$

At α=0.05

p-value < α

We Reject H₀.

The mean value of population is not equal to 20.