Question

A random sample of 10 college students was drawn from a large university. Their ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24 years. We want to determine if we can infer at the 5% significance level that the population mean is not equal to 20.

Answer 1

Answer:

$t=\frac{21.3-20}{\frac{3.199}{\sqrt{10}}}=1.29$    

The degrees of freedom are given by:

$df=n-1=10-1=9$  

And the p value would be:

$p_v =2*P(t_{(9)}>1.29)=0.229$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is different from 20.

Step-by-step explanation:

Information given

22, 17, 27, 20, 23, 19, 24, 18, 19, and 24

The sample mean and deviation for these data are:

$\bar X=21.3$ represent the ample mean

$s=3.199$ represent the sample standard deviation for the sample  

$n=10$ sample size  

$\mu_o =20$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true mean is equal to 20 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 20$  

Alternative hypothesis:$\mu \neq 20$  

The statistic would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

And replacing the info given we got:

$t=\frac{21.3-20}{\frac{3.199}{\sqrt{10}}}=1.29$    

The degrees of freedom are given by:

$df=n-1=10-1=9$  

And the p value would be:

$p_v =2*P(t_{(9)}>1.29)=0.229$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is different from 20.