In a Leichtman Research Group survey of

1100

TV households,

74.5%

of them had at least one Internet-connected TV device (for example, Smart TV, standalone streaming device, connected video game console). A marketing executive wants to convey high penetration of Internet-connected TV devices, so he makes the claim that the percentage of all homes with at least one Internet-connected TV device is equal to

82%.

Test that claim using a

0.01

significance level. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution.

Let p denote the population proportion of all homes with at least one Internet-connected TV device. Identify the null and alternative hypotheses.

H0:

p

equals=

less than<

greater than>

not equals

enter your response here

H1:

p

greater than>

equals=

not equals

less than<

enter your response here

(Type integers or decimals. Do not round.)Part 2Identify the test statistic.

z=enter your response here

(Round to two decimal places as needed.)Part 3Identify the P-value.

P-value=enter your response here

(Round to three decimal places as needed.)Part 4State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Accept

Reject

Fail to reject

the null hypothesis. There

is

is not

sufficient evidence to

support

warrant rejection of

the claim that the percentage of all homes with at least one Internet-connected TV device is equal to

82%.