Question

a poll of 1012 americans showed that 47.4% of the respondents prefer to watch the news rather than read or listen to it. use those results with a 0.01 significance level to test the claim that fewer than half of americans prefer to watch the news rather than read or listen to it. use the p - value method. use the normal distribution as an approximation to the binomial distribution.
let p denote the population proportion of all americans who prefer to watch the news rather than read or listen to it. identify the null and alternative hypotheses.
$h_0: p = 0.5$
$h_1: p < 0.5$
(type integers or decimals. do not round.)
identify the test statistic.
$z = $
(round to two decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=0.474$, $n = 1012$, $p_0=0.5$, $q_0 = 1 - p_0=0.5$

Step2: Compute the test - statistic formula

The formula for the test - statistic in a proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0q_0}{n}}}$
Substitute the values:

$$
LATEXBLOCK0
$$

$\sqrt{1012}\approx31.81$, then

$$
LATEXBLOCK1
$$

Answer:

$z=-1.65$