Question

Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disordered suspects attract more attention from bystanders, they may be more likely to be arrested than nonmentally disordered suspects. Assume that the average number of bystanders when police encounter a suspect is 3.58. In a sample of 20 police encounters with mentally disordered suspects, Engel and Silver (2001) found an average of 7.03 bystanders (standard deviation = 9.42).Perform a hypothesis test to determine if there are more bystanders when the suspect is mentally disordered. Use a 0.01 level of statistical significance. Which distribution, z or t, will you use?

Answer 1

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$    

$p_v =P(t_{(19)}>1.638)=0.0589$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

Step-by-step explanation:

Data given and notation    

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation    

$n=20$ sample size    

$\mu_o =3.58$ represent the value that we want to test  

$\alpha=0.01$ 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 mean is higher than 3.58 :    

Null hypothesis:$\mu \leq 3.58$    

Alternative hypothesis:$\mu > 3.58$    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

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

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$    

P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since is a right tailed test the p value would be:    

$p_v =P(t_{(19)}>1.638)=0.0589$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.