Question

a poll of 1010 americans showed that 47.4% of the respondents prefer to watch the news rather than read or listen to it. use these 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.

identify the test statistic

$h_0:p = 0.5$
$h_1:p < 0.5$
(type integers or decimals. do not round.)

identify the test statistic

$z=-1.66$
(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

This is a left - tailed test since \(H_1:p < 0.5\).

Step2: Use the formula for the P - value in a left - tailed z - test

The formula for the P - value in a left - tailed test is \(P - value=\Phi(z)\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution. Given \(z=-1.66\).
We know that for a standard normal variable \(Z\), we can use a standard normal table or a calculator with a normal distribution function. Using a standard normal table or a calculator (e.g., in Excel: NORM.S.DIST(-1.66, TRUE)), we find that \(P - value=\Phi(-1.66)\).

Step3: Calculate the P - value

Using a standard normal table or a calculator, \(\Phi(-1.66)=0.049\) (rounded to three decimal places).

Answer:

\(0.049\)