Public health officials claim that people living in low income neighborhoods hav

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 1.63 and the standard deviation is 0.52. A study took a random sample of 50 people who lived in low income neighborhoods and found their mean PAL to be 1.53. Using a one-sample z test, what is the z-score for this data? What is the critical value of z in the previous question? Report the critical value CLOSEST to your calculated z-score. (Answer to 2 decimal places.)

Explanation / Answer

Solution:

We will use one sample z test for the population mean.

The formula for z-score is given as below:
Z = (Xbar - µ) / [ / n ]

We are givenPopulation mean = µ = 1.63

Population standard deviation = = 0.52

Sample size = n = 50

Sample mean = Xbar = 1.53

Now, plug all values in the above formula and calculate the value for Z score given as below:

Z = (1.53 – 1.63)/[0.52/50]

Z = -1.35982

Z-score = -1.36 ( two decimal place )

Answer

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