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 1.64 and the standard deviation is 0.53. A study took a random sample of 45 people who lived in low income neighborhoods and found their mean PAL to be 1.59.

A. Using a one-sample z test, what is the z-score for this data? (Answer to 2 decimal places)

B. What is the critical value of z from part A? Report the critical value CLOSEST to your calculated z-score. (Answer to 2 decimal places.)

Please give correct answer and show all work so I can follow along!

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 1.59

H_a :    1.59

= 1.64

= 1.59

= 0.53

n = 45

A)

Test statistic = z

= ( - ) / / n

= (1.64 - 1.59) / 0.53 / 45

= 0.63

Test statistic = 0.63

z score= 0.63

B)

= 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Critical value = +/-1.96

Test statistic < critical value .

Fail to reject the null hypothesis .