Question

a poll of 1161 americans showed that 46.6% of the respondents prefer to watch the news rather than read or listen to 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=square ) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=0.466$, the hypothesized proportion $p = 0.5$, and the sample size $n=1161$.

Step2: Calculate the test - statistic formula

The formula for the test - statistic $z$ in a proportion test is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$.
Substitute the values: $z=\frac{0.466 - 0.5}{\sqrt{\frac{0.5\times(1 - 0.5)}{1161}}}$.
First, calculate the denominator: $\sqrt{\frac{0.5\times0.5}{1161}}=\sqrt{\frac{0.25}{1161}}\approx\sqrt{0.0002153}\approx0.0147$.
Then, calculate the numerator: $0.466−0.5=- 0.034$.
So, $z=\frac{-0.034}{0.0147}\approx - 2.31$.

Answer:

$z=-2.31$