Question

A recent broadcast of a television show had a

1010

share, meaning that among

60006000

monitored households with TV sets in use,

1010 %

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

1515 %

were tuned into the program. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, 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.

A.

H0 :

pequals=0.150.15

H1 :

pless than<0.150.15

B.

H0 :

pequals=0.850.85

H1 :

pnot equals≠0.850.85

C.

H0 :

pequals=0.150.15

H1 :

pgreater than>0.150.15

D.

H0 :

pequals=0.850.85

H1 :

pless than<0.850.85

E.

H0 :

pequals=0.850.85

H1 :

pgreater than>0.850.85

F.

H0 :

pequals=0.150.15

H1 :

pnot equals≠0.150.15

The test statistic is

zequals=nothing.

(Round to two decimal places as needed.)The P-value is

nothing.

(Round to four decimal places as needed.)

Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.

▼

Fail to reject

Reject

H0.

There

▼

is

is not

sufficient evidence to support the claim that less than

1515 %

of the TV sets in use were tuned to the program.

Answer 1

A. H0: p=0.15, H1: p<0.15

We have p_ hat= sample proportion= 0.1

n= sample size= 6000

The test statistic, z= (p_ hat-p)/se

se=√p(1-p)/n= 0.0046

z= (0.1-0.15)/0.0046 = -10.85

p-value= P(z<-10.85) = 0.0000

As p-value is less than alpha= 0.01 so Reject H0.

There is sufficient evidence to support the claim that less than 15% of the TV sets in use were tuned to the program.