Question

10. The stimulant Ritalin has been shown to increase attention span and improve academic performance in children with ADHD (Evans et al., 2001). To demonstrate the effectiveness of the drug, a researcher selects a sample of the disorder and measures each child's attention span before and after taking the drug. The data show an average increase of attention span of Mp = 4.8 minutes with a variance of s2 125 for the sample of difference scores. 20 children diagnosed with a. Is this result sufficient to conclude that Ritalin significantly improves attention span? Use a one-tailed test with a = .05. b. Compute the 80% confidence interval for the mean change in attention span for the population.

Answer 1

The standard deviation is difference is

Si= 125 11.1803

(a)

Here we have d 4.8, s11.1803, n =20 Hypotheses are: Но : д, зо Н.: д, >0 Level of significance: a 0.05 Test is one tailed (right tailed) df =n-1=19 Degree of freedom: Test Statistics: Prd-P (4.8-0)/(11.1803/sqrt(20)) 1.92 Critical value: 1.729 Ift 1.729, Reject HO Rejection Region: reject the null hypothesis Decision: Since test statistics lies in rejection region So we P-value: P-value-0.035 Decision: Since p-value is less than level of significance so reject the null hypothesis O we Excel function for critical value: =TINV(0.1,19) =TDIST (1.92,19,1) Excel function for p-value:

Conclusion: This result is sufficient to conclude that Ritalin significantly improves attention span.

(b)

Since populaiton SD is unknown so t critical value will be used Level of significance: a 0.2 Critical value: t=1.328 df n-1=19 Degree of freedom: The requried confidence interval is: 4.8 (11.1803*1.328/sqrt(20)) 4.8 t 3.32 (1.48,8.12) =TINV (0.2,19) Excel function for critical value: Hence, the required confidence interval is (1.48,8.12).