19) Two methods, A & B, for performing knee-replacement surgery are compared. Me

Question

19) Two methods, A & B, for performing knee-replacement surgery are compared. Method A was used on 40 patients, method B on 50: average recovery time = 32.8 days average recovery time = 30.7 days, SD = 4.6 days SD = 5.4 days A Determine the p-value in a test of the hypothesis that A &B; are equal, on average vs. the alternative that A results in a longer average recovery time. a) .012 b).023c).036 d) .044e).056 20) Continuing (19), a 90% confidence interval for difference between the population average recovery times is 2.1 ± a) 1.73 b) 1.89c) 2.14d) 1.25e) 1.12 21) Repeat problem (19), but with sample sizes of 8 & 10 (assume the observations are from independent normal distributions having equal variances) a) 125 b) .163c).198 ).216 e) .254

Explanation / Answer

21.

Using Minitab

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 8 32.80 4.60 1.6
2 10 30.70 5.40 1.7

Difference = mu (1) - mu (2)
Estimate for difference: 2.10
90% lower bound for difference: -1.11
T-Test of difference = 0 (vs >): T-Value = 0.87 P-Value = 0.198 DF = 16
Both use Pooled StDev = 5.0656

Correct choice is c)0.198

Q.19

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 32.80 4.60 0.73
2 50 30.70 5.40 0.76

Difference = mu (1) - mu (2)
Estimate for difference: 2.10
95% lower bound for difference: 0.35
T-Test of difference = 0 (vs >): T-Value = 1.99 P-Value = 0.025 DF = 87

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 32.80 4.60 0.73
2 50 30.70 5.40 0.76

Difference = mu (1) - mu (2)
Estimate for difference: 2.10
95% upper bound for difference: 3.85
T-Test of difference = 0 (vs <): T-Value = 1.99 P-Value = 0.975 DF = 87

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 32.80 4.60 0.73
2 50 30.70 5.40 0.76

Difference = mu (1) - mu (2)
Estimate for difference: 2.10
95% upper bound for difference: 3.88
T-Test of difference = 0 (vs <): T-Value = 1.96 P-Value = 0.973 DF = 88
Both use Pooled StDev = 5.0611

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 32.80 4.60 0.73
2 50 30.70 5.40 0.76

Difference = mu (1) - mu (2)
Estimate for difference: 2.10
90% lower bound for difference: 0.71
T-Test of difference = 0 (vs >): T-Value = 1.96 P-Value = 0.027 DF = 88
Both use Pooled StDev = 5.0611

P-value is 0.027 its approximately consider 0.023.

Hope this will be helpful. Thanksa nd God Bless yOu :-)

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