Question

Suppose a new standardized test is given to 98 randomly selected third-grade students in New Jersey. The sample average score Y on the test is 57 points, and the sample standard deviation, sy, is 10 points. 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 ( 55.02, 58.98 ). (Round your responses to two decimal places.) Suppose the same test is given to 196 randomly selected third graders from lowa, producing a sample average of 61 points and sample standard deviation of 13 points. The 90% confidence interval for the difference in mean scores between Iowa and New Jersey is C D Round your responses to two decimal places.)

Answer 1

1) Margin of error = $(\frac{1.9847*10}{\sqrt{98}})$

= 2.0049

95% Interval

(57-2.0049, 57+2.0049)

= (55.00, 59.00)

b) Margin of error = $(\frac{1.6527*13}{\sqrt{196}})$

= 1.5347

90% Interval

(61-1.5347, 61+1.5347)

= (59.47, 62.53)