Answer:
Step-by-step explanation:
[tex]H_0: p = 0.5\\H_a : p >0.5[/tex]
(One tailed test at 5% significance level)
Sample size =200 and sample proportion =[tex]\frac{80}{200} =0.40[/tex]
Std error of p = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.4(0.6)}{200} } \\=0.0346[/tex]
Test statistic = p diff/std error = -2.89
p value = 0.001926
So we reject H0 and accept alternate hypothesis.