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 1 and Table 2.
Select the null and the alternative hypotheses to test the policy analyst’s objective.
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. Do not round intermediate calculations. Round \"test statistic\" to 2 decimal places and \"p-value\" to 4 decimal places.)
At = 0.05 has the law been effective?
Test using the critical value approach with = 0.05.
Calculate the critical value. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)
Interpret the results.
| 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 1 and Table 2. | 
Solution
Set Up Hypothesis
 Null, H0:P 0.82
 Alternate, 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
 [ANSWERS]
 a. H0: p  0.82; HA: p < 0.82
 b. Zo =-2.5767 ~ -2.58, ( P < -2.57674 ) = 0.00499
 c. Yes
 d. Critical Value : -1.64
 e. Reject H0

