Question

In a test of physical fitness, a group of men ages 65 and older from a local retirement community were told to do as many sit-ups as they could. It is known the population mean is 20 with a population standard deviation of 8. The scores from the retirement community are given below. Use a two-tailed region of rejection with a criterion probability of .05. What is (answer a-f),

24, 25, 10, 23, 18, 29, 47, 32, 19, 24, 20, 28, 21, 20, 27, 25

a) the sample mean?

b) the standard error of the mean

c) the z-obtained (also referred to as z-score, standardized mean)?

d) what is z-critical value should you use?

e) does the sample represent the population or should you conclude it comes from a different population and why?

f) interpret the results as you would for a reporter - what does this mean in terms of men in the sample compared in the men in the population (use the independent and dependent variable in your interpretation, write these as a statement)?

Answer 1

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The population mean is 20.

Alternative hypothesis: Ha: The population mean is not 20.

H0: µ = 20 versus Ha: µ ≠ 20

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 20

(Part a)

Sample mean = Xbar = 24.5

σ = 8

n = 16

(Part b)

Standard error = σ/sqrt(n) = 8/sqrt(16) = 8/4 = 2

Standard error = 2.00

(Part c)

Z = (24.5 – 20)/[8/sqrt(16)]

Z = 2.25

(Part d)

α = 0.05

Critical value = -1.96 and 1.96

(by using z-table or excel)

Part e

Test statistic is not lies between critical values.

So, we reject the null hypothesis

There is sufficient evidence to conclude that the sample is come from different population.

Part f

There is not sufficient evidence to conclude that the population mean is 20.

We can write the above statement as there is sufficient evidence to conclude that the population mean is not 20.