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 1.61 and the standard deviation is 0.54. A study took a random sample of 45 people who lived in low income neighborhoods and found their mean PAL to be 1.51. Using a one-sample z test, what is the z-score for this data? (Answer to 2 decimal places) 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:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 1.61

Alternative hypothesis: 1.61

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z -test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.0805

z = (x - ) / SE

z = - 1.24

zcritical = + 1.96

zcritical = - 1.96 is closest to z = - 1.24.

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic less than -1.24 or greater than 1.24.

Thus, the P-value = 0.215

Interpret results. Since the P-value (0.215) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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