A survey claims that more than 8 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim, a random sample of 100 doctors’ results in 86 who indicate that they recommend aspirin.
A. 0.0521
B. 0.0802
C. 0.1231
D. 0.0721
E. 0.0668
So far I got:
Ho: P= 0.8
H1: P> 0.8
Z= (0.86-0.8)/0.4 = 1.50
If z= 1.50, cumulative probability is 0.9332
p-value= 2P(Z>zo)
= 2(1-0.9332)=0.1336
Correct answer seems to be 0.0668, which is half of 0.1336. Not sure where I went wrong
$$\begin{split} H_0:\pi=.8\\ H_1:\pi>.8 \end{split} $$
The test statistic is $z^*=\frac{.86-.8}{.4}$
So far so good
You are conducting a one-sided test, so you are only interested in the probability to the right of $z^*$. There is no need to multiply the area by two. The pvalue is is $P(Z>z^*)= 1-0.9332=0.0668$
There is some evidence against the null, that is, it is slightly unlikely that the you would see a sample proportion of .86 if the null hypothesis were true. Therefore it's possible that the true population proportion is greater than .8.