Question

Most adults would not erase all of their personal information online if they could. A software firm survey of 480 randomly selected adults showed that 35​% of them would erase all of their personal information online if they could.Find the value of the test statistic?

Answer 1

Answer:

-7.90

Step-by-step explanation:

Let's write the null hypothesis (H0) and alternative hypothesis (Ha) from the claim statement

P = population proportion of adults who would erase all of their personal information online if they could

H_{0}:P=0.5

H1 P< 0.5

Let's write the given information:

n = sample size = 480

sample proportion = $\hat p = 0.35$

formula of z test statistic for one sample proportion test is as follows:

$Z=\frac{\hat p-P}{\sqrt{\frac{P\times(1-P)}{n}}}$

plug the above values in the above formula we get

$Z =\frac{0.35-0.5}{\sqrt{\frac{0.35(1-0.5)}{480}}}=Z=-7.85584$

So that the test statistic value is -7.90 (Rounded up to two decimal places)