A.)
n = 100 and x = 55
⇒ p = [tex]\frac{55}{100}[/tex] = 0.55
hence p' = 0.50
so, p > 0.50
Z = [tex]\frac{p-p'}{\sqrt{} \frac{p'(1-p')}{n} }[/tex]
Z = [tex]\frac{0.55-0.50}{\sqrt{} \frac{0.50(1-0.50)}{100} }[/tex]
so, Z = 1
let level of significance α = 0.05
critical Z value at α = 0.05 for one tail test is 1.645
since Z = 1 < z' = 1.645
do not reject at α = 0.05
there is not sufficient evidence that more than half of the school student recycle.
B.)
given data,
X = 63 and n = 100
now P = 63/100 = 0.63
Z = [tex]\frac{0.63-0.50}{\sqrt{} \frac{0.50(1-0.50)}{100} }[/tex] = 2.6
since, Z = 2.6 > Z' = 1.645
⇒ reject at α = 0.05
there is sufficient evidence that more than half or majority school student recycle.
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