Question

In a recent Super Bowl, a TV network predicted that 25 % 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 150 people who saw the commercials and found that 30 of them said they would watch one of the television shows. Suppose you have the following null and alternative hypotheses for a test you are running:

H0:p=0.53H0:p=0.53
Ha:p≠0.53Ha:p≠0.53Calculate the test statistic.

Answer 1

The right solution is "-8.25".

Explanation:

The given values are:

Sampled people,

n = 150

Null and alternative hypotheses will be:

`H_0:p=0.53`

`Ha:p\\eq 0.53`

The z-statistics will be:

⇒
`z=\frac{\bar{p}-p_0}{\sqrt{(p_0(1-p_0))/(n) } }`

On substituting the values, we get

⇒
`=\frac{0.2-0.53}{\sqrt{(0.53(1-0.53))/(150) } }`

⇒
`=\frac{-0.33}{\sqrt{(0.24)/(150)} }`

⇒
`=(-0.33)/(0.04)`

⇒
`=-8.25`