Let p be the proportion of the correct answers.
As per given , we have
[tex]H_0: p=0.50[/tex]
[tex]H_a: p>0.50[/tex] , since the alternative hypothesis is right tailed , so the test is a one -tailed test.
If the student gets 52 answers correct out of 80.
i.e. the proportion of correct answers : [tex]\dfrac{52}{80}=0.65[/tex]
Test statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.65-0.50}{\sqrt{\dfrac{0.50(0.50)}{80}}}\approx2.68[/tex]
P-value : [tex]P(z>2.68)=0.0037[/tex] [ by using p-value table for z (right-tailed)]
Since the p-value(0.0037) is less than the significance level (0.05), so we reject the null hypothesis.
Results : We have enough evidence to support the claim that a student knows more than half of the answers and is not just guessing.
