Question

The following table provides summary statistics for the DurationSurgery based on whether or not patients contracted an SSI from the Seasonal Effect data set. One of the researchers is curious whether there is evidence to suggest that surgery duration was longer in patients who contracted SSIs. Use the following information to conduct the following hypothesis test:

• A one-tail T-test for a two-sample difference in means at the 99% confidence level
• with Null Hypothesis that the average surgery duration in patients that did contract SSIs is equal to the average surgery duration in patients that did not contract SSIs
• and with Alternate Hypothesis that the average surgery duration in patients that did contract SSIs is greater than the average surgery duration in patients that did not contract SSIs

Seasonal Effect	Duration of Surgery
	Average	St. Dev.	Count
No SSI	3.506	1.899	2678
Yes SSI	4.418	2.243	241

a. Calculate the standard error of the mean for each group. (10%)

b. Using the correct degrees of freedom (df = group X size + group Y size ̶ # of groups), the correct number of tails, and at the correct confidence level, determine the critical value of t. (10%)

c. Explain under which scenarios using a pooled variance be inadvisable, then, calculate the pooled variance (formula for S² is onpage 379) for the groups. (10%)

d. Calculate the test statistic, T_test (formula for t is on page 380). (10%)

e. The sleep center’s statistician tells you that the p-value for the test is less than 0.0001. Summarize the result of the study. Compare the mean scores in each group. Compare the test statistic to the critical value. Compare the p-value to alpha. Do you find a statistically significant difference? Is there a meaningful/practical difference? Explain your decisions and Justify your claims. (15%)

Answer 1

Claim : the average surgery duration in patients that did contract SSIs is greater than the average surgery duration in patients that did not contract SSIs

Let µ₁ = The average surgery duration in patients that did contract SSIs and

µ₂ = The average surgery duration in patients that did not contract SSIs

H₀: µ₁ = µ₂    vs H_a: µ₁ > µ₂   

a)

Standard error for SSI = Std.dev / = 2.243 /   = 0.1445

Standard error for No SSI = Std.dev / = 1.899/   = 0.0367

b) Critical value t

We are given confidence level = 0.99

α =1 - confidence level = 1- 0.99 = 0.01

degrees of freedom (df) = n₁+ n₂ - 2 = 241+2678-2 = 2917

We have d.f = 2917 , since it is not available on the table , we have to find critical value for d.f = 1000 and α= 0.01 for one tail test.

Therefore critical value = 2.33

c) If population variances of both groups are different σ²₁ ≠ σ²₂  then pooled variance will be inadvisable,

Pooled variance :

S² =

S₁ = 2.243 , n₁ = 241 and S₂= 1.899, n₂ = 2678

S² =

S² = 3.7234

d) Test statistic t =   

We have S² = 3.7234 , therefore S = 1.9296

t =

t = 7.03

e)

Decision rule for p-value : Reject H₀, if p-value < α  Or fail to reject H₀ , if p-value > α  

We have α = 0.01  critical value = 2.33

We are given that p-value for the test is less than 0.0001.

Therefore p-value is also less than α = 0.01 , so we reject the null H₀.

Critical region : Reject H₀, if t ≥ 2.33 Or fail to reject H₀ , if t < 2.33

As t = 7.03 is greater than 2.33 , we reject null H₀.

Conclusion : We have significant evidence that the average surgery duration in patients that did contract SSIs is greater than the average surgery duration in patients that did not contract SSIs  

Effect size ; S_p    = 2.0781

Effect size =   0.44

So effect size 0.44 indicates the difference between two groups is medium.