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
$p - value=square$
(round to three decimal places as needed)

Explanation:

Step1: Determine the type of test

Since \(H_1:p
eq0.78\), this is a two - tailed test.

Step2: Calculate the P - value

For a two - tailed \(z\) - test, the P - value is \(2\times(1 - \Phi(|z|))\). Given \(z=- 2.52\), then \(|z| = 2.52\).
Using the standard normal distribution table or a calculator, \(\Phi(2.52)\) (the cumulative distribution function of the standard normal distribution at \(z = 2.52\)) is approximately \(0.9941\).
So, \(P-\text{value}=2\times(1 - 0.9941)\)
\(=2\times0.0059\)
\(=0.0118\approx0.012\)

Answer:

\(0.012\)