Question

After extensive measurements of the time necessary to complete the first homework assignment, a teacher determines that there is a population mean of 100 and a stardard deviation of 20. If she samples a class of 60 students and calculates a mean of 96 minutes, what is the z statistic (round to the nearest 2 decimal places, don't forget a negative sign if necessary)?

Answer 1

Answer: -1.55

Step-by-step explanation:

Given : Mean : $\mu=100$

Standard deviation : $\sigma=20$

Sample size : $n=60$

Sample mean : $\overline{x}=96$

The test statistic for the population mean is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{96-100}{\dfrac{20}{\sqrt{60}}}=-1.54919333848\approx-1.55$

Hence, the value of z statistic = -1.55