Answer: The value of test statistic is -0.052.
Step-by-step explanation:
Since we have given that
n = 677
and
[tex]\bar {p}=0.499[/tex]
Claim that Most of the adults erase all of their personal details online.
p = 0.5
q = 1-0.5 = 0.5
So, the test statistic would be
[tex]z=\dfrac{\bar{p}-p}{\sqrt{\dfrac{pq}{n}}}\\\\z=\dfrac{0.499-0.5}{\sqrt{\dfrac{0.5\times 0.5}{677}}}\\\\z=\dfrac{-0.001}{0.0192}\\\\z=-0.052[/tex]