Suppose a new standardized test is given to 99 randomly selected third-grade students in New Jersey. The sample average score Upper Y overbar on the test is 55 points, and the sample standard deviation, s Subscript Upper Y, is 10 points.

a. The authors plan to administer the test to all third-grade students in New Jersey. The 95% confidence interval for the mean score of all New Jersey third graders is ( 53.0204, 56.9796). (Round your responses to two decimal places.)
b. Suppose the same test is given to 198 randomly selected third graders from Iowa, producing a sample average of 59 points and sample standard deviation of 13 points. The 90% confidence interval for the difference in mean scores between Iowa and New Jersey is ( nothing, nothing). (Round your responses to two decimal places.)

(a) 95% confidence interval for the mean score of all New Jersey third graders is (53.01, 56.99)

(b) 90% confidence interval for the difference in mean scores between lowa and New Jersey is (3.71, 4.29)

Explanation:

Confidence Interval = mean + or - Error margin (E)

(a) New Jersey

mean = 55

sd = 10

n = 99

degree of freedom = n - 1 = 99 - 1 = 98

Confidence level (C) = 95% = 0.95

Significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

t-value corresponding to 98 degrees of freedom and 5% significance level is 1.9846

E = t×sd/√n = 1.9846×10/√99 = 1.99

Lower limit = mean - E = 55 - 1.99 = 53.01

Upper limit = mean + E = 55 + 1.99 = (56.99)

95% confidence interval is between 53.01 and 56.99.

(b) Lowa

mean = 59

sd = 13

n = 198

Difference in mean between Lowa and New Jersey = 59 - 55 = 4

Difference in sd between Lowa and New Jersey = 13 - 10 = 3

degree of freedom = n1 + n2 - 2 = 99 + 198 - 2 = 295

Confidence level = 90%

Significance level = 10%

t-value corresponding to 295 degrees of freedom and 10% confidence interval is 1.65312

E = t×sd/√n = 1.65312×3/√297 = 0.29

Lower limit = mean - E = 4 - 0.29 = 3.71

Upper limit = mean + E = 4 + 0.29 = 4.29

90% confidence interval is between 3.71 and 4.29

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