Question

Public health officials claim that people living in low income neighborhoods have different Physical Activity Levels (PAL) than the general population. This is based on knowledge that in the U.S., the mean PAL is 1.6 and the standard deviation is 0.45. A study took a random sample of 44 people who lived in low income neighborhoods and found their mean PAL to be 1.56. Using a one-sample z test, what is the z-score for this data? (Answer to 2 decimal places)

Answer 1

Answer: -0.59

Step-by-step explanation:

Given : Population mean : $\mu=1.6$

Population standard deviation : $\sigma=0.45$

Sample size : n= 44

Sample mean : $\overline{x}=1.56$

One-step z-test for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\=\dfrac{1.56-1.6}{\dfrac{0.45}{\sqrt{44}}}\\\\=-0.589622184952\approx-0.59$

Hence, the z-score for this data =-0.59