in a recent super bowl, a tv network predicted that 51% of the audience would express an interest in seeing…

in a recent super bowl, a tv network predicted that 51% of the audience would express an interest in seeing one of its forthcoming television shows. the network ran commercials for these shows during the super bowl. the day after the super bowl, and advertising group sampled 70 people who saw the commercials and found that 37 of them said they would watch one of the television shows. suppose you are have the following null and alternative hypotheses for a test you are running: $h_0:p = 0.51$ $h_a:p\neq0.51$ calculate the test statistic, rounded to 3 decimal places $z=$
Answer
Explanation:
Step1: Calculate sample proportion
The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 37$ (number of successes) and $n=70$ (sample - size). So, $\hat{p}=\frac{37}{70}\approx0.5286$.
Step2: Identify hypothesized proportion
The hypothesized proportion $p_0 = 0.51$.
Step3: Calculate the test - statistic formula
The formula for the z - test statistic in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Step4: Substitute values into the formula
Substitute $\hat{p}=0.5286$, $p_0 = 0.51$, and $n = 70$ into the formula: [ \begin{align*} z&=\frac{0.5286 - 0.51}{\sqrt{\frac{0.51\times(1 - 0.51)}{70}}}\ &=\frac{0.0186}{\sqrt{\frac{0.51\times0.49}{70}}}\ &=\frac{0.0186}{\sqrt{\frac{0.2499}{70}}}\ &=\frac{0.0186}{\sqrt{0.00357}}\ &=\frac{0.0186}{0.05975}\ &\approx0.311 \end{align*} ]
Answer:
$0.311$