Respuesta :
Answer:
The p-value of the test is of 0.3219 > 0.05, which means that at [tex]\alpha = 0.05[/tex], the average age is not higher than originally believed.
Step-by-step explanation:
A recent study in a small city stated that the average age of robbery victims was 63.5 years. Test if the average age higher than originally believed.
At the null hypothesis, we test that the mean is the believed mean age of 63.5 years, that is:
[tex]H_0: \mu = 63.5[/tex]
At the alternate hypothesis, we test if this mean is higher, that is:
[tex]H_1: \mu > 63.5[/tex]
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
63.5 is tested at the null hypothesis:
This means that [tex]\mu = 63.5[/tex]
A random sample of 20 recent victims had a mean of 63.7 years and a standard deviation of 1.9 years.
This means that [tex]n = 20, X = 63.7, s = 1.9[/tex]
Test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{63.7 - 63.5}{\frac{1.9}{\sqrt{20}}}[/tex]
[tex]t = 0.47[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 63.7, which is a right-tailed test, with t = 0.47 and 20 - 1 = 19 degrees of freedom.
With the help of a calculator, this probability is of 0.3219.
The p-value of the test is of 0.3219 > 0.05, which means that at [tex]\alpha = 0.05[/tex], the average age is not higher than originally believed.
