Question

Analysis of methods of teaching reading to “slow learners” is conducted in which a new method is compared to the standard method. Of a random sample of 22 slow learners in the third grade selected in Holland, MI, 10 are taught by the new method and 12 are taught by the standard method. All 22 children are taught by qualified instructors under similar conditions for a 6-month period. At the end of the 6-months, all of the children are given the same reading test and the scores are recorded. The scores were entered in SPSS and the output is below.

A. Is there sufficient evidence to indicate that the new method produces better results than the standard method? Test at α = 0.02. Show all steps.

B. Interpret the 98% confidence interval for the true mean difference in test scores between the new method and the standard method for Holland third graders. Note: ONLY interpret!

Answer 1

(a) The hypothesis being tested is:

H0: µ1 = µ2

H1: µ1 < µ2

The p-value is 0.0079.

Since the p-value (0.0079) is less than the significance level (0.02), we can reject the null hypothesis.

Therefore, we can conclude that the new method produces better results than the standard method.

(b) We are 98% confident that the true mean difference in test scores between the new method and the standard method for Holland third graders is between -7.840 and -0.160.

1	2
12	16	mean
4	5	std. dev.
22	12	n
	32	df
	-4.000	difference (1 - 2)
	19.094	pooled variance
	4.370	pooled std. dev.
	1.568	standard error of difference
	0	hypothesized difference
	-2.551	t
	.0079	p-value (one-tailed, lower)
	-7.840	confidence interval 98.% lower
	-0.160	confidence interval 98.% upper
	3.840	margin of error