1)H0:p=0.3,H1:p=0.3 (2-tailed test)
pN=(0.3)175=5.25>5(1−p)N=(0.7)175>5
So we will use z-test.
z−value=1.96 (α=0.05)
If test statistic >1.96 or <−1.96 we shall reject H0. Otherwise we accept H0.
z=Np(1−p)p∗−p=175(0.3)(0.7)17563−0.3≈1.732 — test statistic.
So we accept H0.
The claim of the cell phone company is not reasonable.
2) H0:a=a0=4,H1:a<a0=4.
We shall use a random variable T=s(x−a0)n and t-test;
k=n−1=36−1=35 — degree of freedom.
tcr=tcr(α;k)=tcr(0.02;35)≈2.1332 (one-sided).tobs=1.21(3.74−4)36≈−1.289.tobs=−1.289>−tcr=−2.1332.
So tobs is not in the critical region (−∞,−2.1332) and we accept H0.
Association's claim is not reasonable.