In a Leichtman Research Group survey of

850

TV households,

75.3%

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

79%.

Test that claim using a

0.10

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

Question content area bottom

Part 1

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

less than<

not equals

greater than>

equals=

enter your response here

H1:

p

less than<

not equals

equals=

greater than>

enter your response here

(Type integers or decimals. Do not round.)

Part 2

Identify the test statistic.

z=enter your response here

(Round to two decimal places as needed.)

Part 3

Identify the P-value.

P-value=enter your response here

(Round to three decimal places as needed.)

Part 4

State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Accept

Fail to reject

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

79%