Step-by-step explanation:
1. the claim here is that over 60 percents of the households living in ann harbor have pets.
null hypothesis;
h0: p≤0.60
alternative:
h1: p>0.60
2.
n is the size of the sample = 280
n(p) = 280 x 0.60 = 168 greater than 10
n(1-p) = 280(1-0.6) = 280 x 0.4 = 112 greater than 10
we use the normal dist since each of the above are greater than 10
3. n = 280
x = 182
x/n = 182/280
= 0.65
we get the test statistic z
z = [tex]\frac{0.65-0.6}{\sqrt{0.6*(1-0.6)/280} }[/tex]
= [tex]\frac{0.05}{\sqrt{0.000857} }[/tex]
=[tex]\frac{0.05}{0.02927}[/tex]
≈ [tex]1.71[/tex]
we get the p value = 1-p(z<1.71)
= 1-.9562
this gives 0.0438
4. we can see that 0.0438 < 0.10
so we conclude that over 60% of these households have at least one pet since we rejected h0.
