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

Respuesta :

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 = ∑[tex]\frac{x_{i} }{N}[/tex], where:

x is sample mean;

[tex]x_{i}[/tex] is each of the values;

N is the quantity the sample has;

From the given data, N = 16, so,

x = [tex]\frac{24+25+10+23+18+29+47+32+19+24+20+28=21+20+27+25}{16}[/tex]

x = 24.5

The sample mean is x = 24.5

Answer:

The sample mean = 24.5

Step-by-step 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:

[tex]z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

The population mean,  µ = 20

The sample mean, [tex]\bar{x} = \frac{\sum x}{n}[/tex]

n = 16

[tex]\bar {x} = \frac{24 + 25 + 10 + 23 + 18 + 29 + 47 + 32 + 19+24+20+28+21+20+27+25}{16}[/tex]

[tex]\bar {x} = \frac{392}{16} \\\bar {x} = 24.5[/tex]