Answer:
99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].
Step-by-step explanation:
We are given that a high school principal wishes to estimate how well his students are doing in math. 
Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.
Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;
                           P.Q. =  ~ N(0,1)
  ~ N(0,1)
where,  = sample proportion of students received a passing grade = 77%
 = sample proportion of students received a passing grade = 77%
            n = sample of tests = 40
            p = population proportion
<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>
So, 99% confidence interval for the population proportion, p is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%
                                            level of significance are -2.5758 & 2.5758}  
P(-2.5758 <  < 2.5758) = 0.99
 < 2.5758) = 0.99
P(  <
 <  <
 <  ) = 0.99
 ) = 0.99
P(  < p <
 < p <  ) = 0.99
 ) = 0.99
<u>99% confidence interval for p</u> = [ ,
 ,  ]
]
 = [  ,
 ,  ]
 ]
 = [0.5986 , 0.9414]
Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].
Lower bound of interval = 0.5986
Upper bound of interval = 0.9414