Question

a poll of 1161 americans showed that 46.6% of the respondents prefer to watch the news rather than read or listen to it. use those results with a 0.10 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=-2.31 ) (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 distribution type

Since we are using the normal distribution as an approximation to the binomial distribution and testing a proportion, we use the standard normal distribution ($Z -$ distribution) for finding the $P -$ value. The $P -$ value for a left - tailed test (because $H_1:p\lt0.5$) is $P(Z\lt z)$, where $z$ is the test statistic.

Step2: Calculate the $P -$ value

We know that $z=-2.31$. Using a standard normal table (or a calculator with a normal distribution function, e.g., in Excel: =NORM.S.DIST(-2.31,TRUE)), we find the probability that a standard normal random variable $Z$ is less than $-2.31$.
From the standard normal table, $P(Z\lt - 2.31)=0.0104$. Rounding to three decimal places.

Answer:

$0.010$