A large university is interested in the outcome of a course standardization process. They have taken a random sample of 100 student grades, across all instructors. The grades represent the proportion of problems answered correctly on a midterm exam. The sample proportion correct was calculated as 0.78. a. Construct a 90% confidence interval on the population proportion of correctly answered problems. b. Construct a 95% confidence interval on the population proportion of correctly answered problems.

Step-by-step explanation:

a)

Number of sample, n = 100

p = 0.78

q = 1 - p = 1 - 0.78

q = 0.22

For a confidence level of 90%, the corresponding z value is 1.645.

The formula for determining the error bound for the proportion is

z × √pq/n

= 1.645 ×√(0.78 × 0.22)/100

= 1.645 × 0.0414 = 0.068

The upper boundary of the population proportion is

0.78 + 0.068 = 0.85

The lower boundary of the population proportion is

0.78 - 0.068 = 0.71

b) For a confidence level of 95%, the corresponding z value is 1.96.

Therefore,

= 1.96 ×√(0.78 × 0.22)/100

= 1.96 × 0.0414 = 0.081

The upper boundary of the population proportion is

0.78 + 0.081 = 0.86

The lower boundary of the population proportion is

0.78 - 0.081 = 0.7

answer: it is 1: 47 pm

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