Answer:
Given:
Sample size, n = 640
Since result showed that 99.2% of them would erase all of their personal information online if they could, we have:
P' = 99.2% = 0.992
Let's take p = 50%
Our null and alternative hypothesis will be :
H0 : u = 0.5
H1 : u > 0.5
In this case we are to check for the test statistic.
[tex] \frac{P' - p}{\sqrt{p(1-p)/n}} [/tex]
[tex] = \frac{0.992 - 50}{\sqrt{50(1-50)/640}} [/tex]
= 24.9
Test statistic, Z = 24.893
In conclusion, the results are significantly high, which suggets there is suffcient evidence to support the claim that most adults would erase all of their personal information online.
