Sample N Mean StDev SEMean 1 60 51.6 23.7 3.1 2 59 51.4 23.2 3.0 Difference = mu

Question

Sample N Mean StDev SEMean

1 60 51.6 23.7 3.1

2 59 51.4 23.2 3.0

Difference = mu(1) - mu(2)
Estimate for difference: 0.200000
95% CI for difference: (-8.316130,8.716130) T-Test of difference = 0(vs not = ):
T@Value = 0.05 P@Value = 0.963 DF = 117
Both use Pooled StDev = 23.4535

Surgery verses placebo for knee pain Refer to Example 10, “Arthroscopic Surgery.” Here we show MINITAB output comparing mean knee pain scores for the placebo (Group 1) to debridement arthroscopic surgery (Group 2).

State and interpret the result of the confidence interval.

State all steps and interpret the result of the significance test.

Based on the confidence interval and test, would you conclude that the arthroscopic surgery works better than placebo? Explain.

Explanation / Answer

state and interpret the result of the confidence interval

95% CI for difference: (-8.316130,8.716130)

- there is 95 % probability that difference between two populaiton mean will lie in this confidence interval

State all steps and interpret the result of the significance test.

Ts = (X1bar - X2bar)/(sp*sqrt(1/n1+1/n2)

=(51.6-51.4)/(23.4535*sqrt(1/60+1/59))

=0.0465

p-value = 0.963

since p-value is more than 0.05

we fail to reject then null hypothesis

Based on the confidence interval and test, would you conclude that the arthroscopic surgery works better than placebo?

similarly as 0 is in confidence interval ,we fail to reject the null hypothesis

and conclude that there is no significant evidence that

that the arthroscopic surgery works better than placebo

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