Question

An SRS of 430 high school seniors gained an average of ¯x=20.44 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 52.32. We want to estimate the mean change in score u in the population of all high school seniors, give a 95% confidence interval (a, b) for based on this sample.

Answer 1

Answer: ( 15.49,25.39)

Therefore at 95% confidence interval (a,b) = ( 15.49,25.39)

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain _x = 20.44

Standard deviation r = 52.32

Number of samples n = 430

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

+20.44+/-1.96(52.32/√430)

+20.44+/-1.96(2.523)

+20.44+/-4.95

= ( 15.49,25.39)

Therefore at 95% confidence interval (a,b) = ( 15.49,25.39)