A recent study by Allstate Insurance Co finds that 82 of tee

A recent study by Allstate Insurance Co. finds that 82% of teenagers have used cell phones while driving (The Wall Street Journal, May 5, 2010). In October 2010, Massachusetts enacted a law that forbids cell phone use by drivers under the age of 18. A policy analyst would like to determine whether the law has decreased the proportion of drivers under the age of 18 who use a cell phone. Use Table 2. Click here to find Table 2 or select the table from D2L

a. Select the null and the alternative hypotheses to test the policy analyst’s objective.

A. Ho: p = 0.82; HA: p 0.82

B. Ho: p 0.82; HA: p > 0.82

C. Ho: p 0.82; HA: p < 0.82

b. Suppose a sample of 200 drivers under the age of 18 results in 150 who still use a cell phone while driving. What is the value of the test statistic? What is the p-value? (Negative values should be indicated by a minus sign. Round your intermediate calculations to 4 decimal places. Round \"test statistic\" to 2 decimal places and \"p-value\" to 4 decimal places.)

c. At = 0.05 has the law been effective?

A. Yes

B. No

Test using the critical value approach with = 0.05.

d-1. Calculate the critical value. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

d-2. What is the conclusion?

A. Reject Ho

B. Do not reject Ho

Solution

Set Up Hypothesis
Null,82% of teenagers have used cell phones while driving H0:P<=0.82
Alternate,decreased the proportion of drivers use a cell phone H1: P>0.82
Test Statistic
No. Of Success chances Observed (x)=150
Number of objects in a sample provided(n)=200
No. Of Success Rate ( P )= x/n = 0.75
Success Probability ( Po )=0.82
Failure Probability ( Qo) = 0.18
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.75-0.82/(Sqrt(0.1476)/200)
Zo =-2.5767
| Zo | =2.5767
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =2.577 & | Z | =1.64
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Left Tail -Ha : ( P < -2.57674 ) = 0.00499
Hence Value of P0.05 > 0.00499,Here we Reject Ho

ANS:
a. Select the null and the alternative hypotheses to test the policy analyst’s objective
B. Ho: p 0.82; HA: p > 0.82

c. At = 0.05 has the law been effective?
A. Yes

Critical value + 1.64
Reject Ho