A researcher intends to estimate the effect of a drug on the scores of human sub

Question

A researcher intends to estimate the effect of a drug on the scores of human subjects performing a task of psychomotor coordination. The members of a random sample of 9 subjects were given the drug prior to testing. The mean score in this group was 9.02, and the sample variance was 17.44. An independent random sample of 10 subjects was used as a control group and given a placebo prior to testing. The mean score in this control group was 15.39, and the sample variance was 27.59. Assuming that the population distributions are normal with equal variances, find a 99% confidence interval for the difference between the population mean scores. The confidence interval is

Explanation / Answer

Confidence Interval:

(x1 bar –x2 bar) (-/+) E

X1 bar= 9.02

X2 bar = 15.39

E = tc * SE

SE = ((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2) * (1/n1+1/n2)

n1 = 9

n2 = 10

s1^2 = 17.44

s2^2 = 27.59

df = n1+n2 – 2 = 9 + 10 – 2 = 17

Level of significance is .01

From the t distribution table at .01 level of significance for 17 degrees of freedom is given by 2.898

SE = ((9-1)*17.44+(10-1)*27.59)/(9+10-2) * (1/9+1/10)

    = 2.1946

E = 2.898 * 2.1964

= 6.3652

X1 bar – x2 bar = 9.02 – 15.39

                            = -6.37

-6.37 (-/+) 6.3652

-12.74 and 0.00

Answer:

-12.74 and 0.00

If we use excel for solution then we get it as

-12.73 and 0.01

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