Question

At a local university, a sample of 49 evening students was selected in order to determine whether the average age of the evening students is significantly different from 21. The average age of the students in the sample was 23 years. The population standard deviation is known to be 3.5 years. Determine whether or not the average age of the evening students is significantly different from 21. Use a 0.1 level of significance.

Answer 1

Answer:

$z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4$    

$p_v =2*P(z>4)=0.0000633$  

When we compare the significance level $\alpha=0.1$ we see that $p_v$ so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

Step-by-step explanation:

Data given and notation  

$\bar X=23$ represent the sample mean

$\sigma=3.5$ represent the population standard deviation

$n=49$ sample size  

$\mu_o =21$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the average age of the evening students is significantly different from 21, the system of hypothesis would be:  

Null hypothesis:$\mu = 21$  

Alternative hypothesis:$\mu \neq 21$  

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z>4)=0.0000633$  

Conclusion  

When we compare the significance level $\alpha=0.1$ we see that $p_v$ so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.