Question

in a leichtman research group survey of 1000 tv households, 74.7% 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 78%. 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
$h_0:p = 0.78$
$h_1:p
eq0.78$
(type integers or decimals. do not round.)
identify the test statistic
$z=-2.52$
(round to two decimal places as needed.)
identify the p - value
$p - value = 0.012$
(round to three decimal places as needed.)
state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim

the null hypothesis. there sufficient evidence to the claim that the percentage of all homes with at least one internet - connected tv device is equal to 78%.

Explanation:

Step1: Compare P - value and significance level

Given significance level $\alpha = 0.01$. The P - value is $0.012$.
Since $P - value=0.012>0.01=\alpha$.

Step2: Make decision about null hypothesis

When $P - value>\alpha$, we fail to reject the null hypothesis.

Answer:

Fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the percentage of all homes with at least one Internet - connected TV device is equal to 78%.