Question

A researcher wants to test the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds x=20.6 months and s=8 months. Test the claim that u=18.7 months at the 0.05 significance level.

Answer 1

Answer:

$t=\frac{20.6-18.7}{\frac{8}{\sqrt{11}}}=0.788$  

The degrees of freedom are given by;

$df =n-1= 11-1=10$

And the p value would be:

$p_v =2*P(t_{10}>0.788)=0.449$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different than 18.7

Step-by-step explanation:

Information given

$\bar X=20.6$ represent the sample mean

$s=8$ represent the sample standard deviation

$n=11$ sample size  

$\mu_o =18.7$ represent the value to test

$\alpha=0.05$ represent the significance level

t would represent the statistic  

$p_v$ represent the p value

Hypotesis to test

We want to verify if the true mean is equal to 18.7, the system of hypothesis would be:  

Null hypothesis:$\mu =18.7$  

Alternative hypothesis:$\mu \neq 18.7$  

The statistic is given by:

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

Replacing the info we got:

$t=\frac{20.6-18.7}{\frac{8}{\sqrt{11}}}=0.788$  

The degrees of freedom are given by;

$df =n-1= 11-1=10$

And the p value would be:

$p_v =2*P(t_{10}>0.788)=0.449$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different than 18.7