A recent broadcast of a television show had a 10 share meani

A recent broadcast of a television show had a 10 share, meaning that among 6000 monitored households with TV sets in use, 10% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in use, less than 15% were tuned into the program. Identify the null hypothesis, alternative hypothesis, test statistic, P-vale, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below. The test statistic is z = (Round to two decimal places as needed.) The P-value is . (Round to four decimal places as needed.) Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim. H0. There sufficient evidence to support the claim that less than 15% of the TV sets in use were tuned to the program.

Solution

Set Up Hypothesis
Null,it is as equal to 15% H0:P=0.15
Alternate, less than 15% are tuned into the program H1: P<0.15
Test Statistic
Number of objects in a sample provided(n)=6000
No. Of Success Rate ( P )= x/n = 0.1
Success Probability ( Po )=0.15
Failure Probability ( Qo) = 0.85
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.1-0.15/(Sqrt(0.1275)/6000)
Zo =-10.8465
| Zo | =10.8465
Critical Value
The Value of |Z | at LOS 0.01% is 2.33
We got |Zo| =10.847 & | Z | =2.33
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Left Tail -Ha : ( P < -10.84652 ) = 0
Hence Value of P0.01 > 0,Here we Reject Ho

ANS. H0:P=0.15 , H0:P<0.15
Zo =-10.85
0
Reject Ho, There is suffcient evidence