Question

An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty years of age. The investigator administers a vocabulary test to a group of 31 younger subjects and to a group of 31 older subjects. Higher scores reflect better performance. The mean score for younger subjects was 14.0 and the standard deviation of younger subject's scores was 5.0. The mean score for older subjects was 20.0 and the standard deviation of older subject's scores was 6.0. Does this experiment provide evidence for the investigator's theory?

Answer 1

Answer:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$  

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}<-4.28)=6.83x10^{-5}$  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

Step-by-step explanation:

Information given

$\bar X_{1}=14$ represent the mean for sample 1 (younger)

$\bar X_{2}=20$ represent the mean for sample 2 (older)  

$s_{1}=5$ represent the sample standard deviation for 1  

$s_{f}=6$ represent the sample standard deviation for 2  

$n_{1}=31$ sample size for the group 2  

$n_{2}=31$ sample size for the group 2  

t would represent the statistic

System of hypothesis

We want to test if  that people under the age of forty have vocabularies that are different than those of people over sixty years of age, the system of hypothesis are:

Null hypothesis:$\mu_{1}-\mu_{2}=0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

The statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=31+31-2=60$  

Replacing the info given we got:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$  

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}<-4.28)=6.83x10^{-5}$  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.