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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)
where, = 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
P( <
<
) = 0.99
P( < p <
) = 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