Answer:
Here we have given two catogaries as degree holder and non degree holder. 
 
So here we have to test the hypothesis that, 
 
H0 : p1 = p2 Vs H1 : p1 not= p2 
 
where p1 is population proportion of degree holder. 
 
p2 is population proportion of non degree holder. 
 
Assume alpha = level of significance = 5% = 0.05 
 
The test is two tailed. 
 
Here test statistic follows standard normal distribution. 
 
The test statistic is, 
 
Z = (p1^ - p2^) / SE 
 
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2] 
 
p1^ = x1/n1 
 
p2^ = x2/n2 
 
p^ = (x1+x2) / (n1+n2) 
 
This we can done in TI_83 calculator. 
 
steps : 
 
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER 
 
Test statistic Z = 1.60 
 
P-value = 0.1090 
 
P-value > alpha 
 
Fail to reject H0 or accept H0 at 5% level of significance. 
 
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.