Answer:
C. 0.0015
Explanation:
Null hypothesis,
![H_0:P_H=P_c](https://img.qammunity.org/2020/formulas/mathematics/college/4fz9nvefnsm8omqojbjhvml3mgkcfyd5aa.png)
Alternative hypothesis,
![H_a:P_H>P_c](https://img.qammunity.org/2020/formulas/mathematics/college/nzgxcp3c7e614yyg87s8o7pvjlp9ltq3u0.png)
Sample proportions:
![P_H=(18)/(20+18)=0.473684\\\\P_c=(6)/(32+6)=0.157895](https://img.qammunity.org/2020/formulas/mathematics/college/tsmcz31m8yejlh1j0mvdanx1y3ljl5dxoe.png)
Pooled proportion,
![p=(x+y)/(n_1+n_2)=(18+6)/(76)=0.315789](https://img.qammunity.org/2020/formulas/mathematics/college/jc3t0ow9dbzhnbudph68c7olyudw95te0s.png)
Under
, the test statistic is given by,
![z_0=\frac{P_A-P_C}{\sqrt{P(1-P)((1)/(n_1)+(1)/(n_2))}}\\\\==\frac{0.473684-0.157895}{\sqrt{0.315789(1-0.315789)((1)/(38)+(1)/(38))}}=2.96](https://img.qammunity.org/2020/formulas/mathematics/college/67x7sjc7k2rsb7hhob6n1b6b8xayu8z8xr.png)
Since the alternative hypothesis is two tailed, the p-value is given by,
p-value= P(z > 2.96)=1 - P(Z ≤ 2.96)=1 - 0.9985
(FROM Z TABLE)
P-value = 0.0015