A government agency was interested in whether wearing a seat

A government agency was interested in whether wearing a seat belt is effective in saving a front - seat occupant’s life in a car crash. A simple random sample of front - seat occupants involved in car crashes is obtained . Among 2823 occupants not wearing seat belts, 31 were fatalities ; and among 7765 occupants wearing seat belts, 16 were fatalities . Let p 1 be the proportion of occupants not wearing seat belts that were fatalit ies , and p 2 be the proportion of occupants wearing seat belts that were fatalities . Test the claim that not wearing seat belts leads to more fatalities.

(a)

Formulate the hypotheses.

(b)

Calculate the value of the test statistic using a pooled sample proportion.

(c)

Calculate the p

-

value using (b).

(d)

Make a decision on the

hypothesis using the p

-

value

from (

c

) at a significance level of 0.01.

(e)

Construct a 99% confidence interval using separated sample proportions.

Solution

a)

Formulating the hypotheses          
Ho: p1 - p2   <=   0  
Ha: p1 - p2   >   0  

b)

Here, we see that pdo =    0   , the
hypothesized population proportion difference.  
          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.010981226      
p2 = x2/n2 =    0.002060528      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.00202781      
          
Thus,          
          
z = [p1 - p2 - pdo]/sd =    4.399179194   [ANSWER]

*****************

c)  
          
As significance level =    0.01   , then the critical z is  
          
zcrit =    2.326347874      
          
Also, the p value is          
          
P =    5.43305*10^-6   [ANSWER]  

*********************

d)

As P < 0.01, we Reject the Null Hypothesis.

*****************

e)          

NOT WEARING SEATBELTS:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.010981226          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.001961427          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.005052302          
lower bound = p^ - z(alpha/2) * sp =   0.005928923          
upper bound = p^ + z(alpha/2) * sp =    0.016033528          
              
Thus, the confidence interval is              
              
(   0.005928923   ,   0.016033528   ) [ANSWER]

*********

WEARING SEATBELTS:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.002060528          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.000514601          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.001325524          
lower bound = p^ - z(alpha/2) * sp =   0.000735004          
upper bound = p^ + z(alpha/2) * sp =    0.003386052          
              
Thus, the confidence interval is              
              
(   0.000735004   ,   0.003386052   ) [ANSWER]