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 1.65 and the standard deviation is 0.55. A study took a random sample of 51 people who lived in low income neighborhoods and found their mean PAL to be 1.63. Using a one-sample z test, what is the z-score for this data

Answer 1

Answer:

The z-score for this data is Z = -0.26.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

This is based on knowledge that in the U.S., the mean PAL is 1.65 and the standard deviation is 0.55.

This means that $\mu = 1.65, \sigma = 0.55$

A study took a random sample of 51 people who lived in low income neighborhoods and found their mean PAL to be 1.63.

This means that $n = 51, X = 1.63$

Using a one-sample z test, what is the z-score for this data

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1.63 - 1.65}{\frac{0.55}{\sqrt{51}}}$

$Z = -0.26$

The z-score for this data is Z = -0.26.