Question

More than 1 million high school students took the SAT in 2000. The average verbal score was 428 and the standard deviation was 110.

(a) Estimate the 60th percentile of the verbal SAT scores in 2000.

(b) In California, the average verbal score was 417 and the standard deviation was 110 points. About what percent of the California test takers did better than the national average?

(c) Estimate the difference between national and California’s and national SAT scores. Construct a 95% confidence interval of the difference.

Answer 1

µ = 428

σ = 110

(a) The z-score for p-value = 0.60 is 0.25.

z = (x - µ)/σ

0.25 = (x - 428)/110

x = 455.8682

(b) x = 417

z = (x - µ)/σ

z = (417 - 428)/110

z = -0.1

The probability is 0.5398.

(c)

1	2
428	417	mean
110	110	std. dev.
30	30	n
	11.000	difference (1 - 2)
	28.402	standard error of difference
	0	hypothesized difference
	0.39	z
	.6985	p-value (two-tailed)
	-44.667	confidence interval 95.% lower
	66.667	confidence interval 95.% upper
	55.667	margin of error

The difference between national and California’s and national SAT scores is 11.
The 95% confidence interval of the difference is between -44.667 and 66.667.