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 meanis 20 with The scores for the men from the retirement community are given below. Use a two-tailed rejection region with a total area of 0.05. What is the sample mean?

24 25 10 23 18 29 47 32

19 24 20 28 21 20 27 25

Answer 1

The sample mean = 24.5

Explanation:

Solution:

Using the one sample z test

Since the test is a two tailed test

Null hypothesis, H₀: μ = 20.

Alternative hypothesis, Ha: µ ≠ 20

The test statistic formula is given by:

`z = \frac{\bar{x} - \mu}{(\sigma)/(√(n) ) }`

The population mean, µ = 20

The sample mean,
`\bar{x} = (\sum x)/(n)`

n = 16

`\bar {x} = (24 + 25 + 10 + 23 + 18 + 29 + 47 + 32 + 19+24+20+28+21+20+27+25)/(16)`

`\bar {x} = (392)/(16) \\\bar {x} = 24.5`

Answer 2

Answer: x = 24.5

Step-by-step explanation: Sample mean is the average value of a fraction of the whole population. It is also an estimated of how or what the whole population is doing. To determine it, there is the following formula:

x = ∑
`(x_(i) )/(N)`, where:

x is sample mean;

`x_(i)` is each of the values;

N is the quantity the sample has;

From the given data, N = 16, so,

x =
`(24+25+10+23+18+29+47+32+19+24+20+28=21+20+27+25)/(16)`

x = 24.5

The sample mean is x = 24.5