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.
$z=square$
(round to two decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is given as $0.747$ (since $74.7\% = 0.747$), $n = 1000$, and $p_0=0.78$.

Step2: Calculate the test - statistic formula

The formula for the test - statistic $z$ in a proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

$-2.52$