Question

a poll of 1019 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:plt0.5 ) (type integers or decimals. do not round.) identify the test statistic. ( z=square ) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}$ is given as $0.474$.

Step2: Calculate the standard deviation $\sigma_{\hat{p}}$

The formula for the standard deviation of a proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Here, $p = 0.5$ (from the null hypothesis), $n=1019$.
So, $\sigma_{\hat{p}}=\sqrt{\frac{0.5\times(1 - 0.5)}{1019}}=\sqrt{\frac{0.25}{1019}}\approx0.0157$.

Step3: Calculate the test statistic $z$

The formula for the $z$ - statistic in a proportion test is $z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}$.
Substitute $\hat{p}=0.474$, $p = 0.5$, and $\sigma_{\hat{p}}\approx0.0157$ into the formula:
$z=\frac{0.474 - 0.5}{0.0157}=\frac{- 0.026}{0.0157}\approx - 1.66$.

Answer:

$z\approx - 1.66$