Question

Several neurosurgeons wanted to determine whether a dynamic system (Z-plate) reduced operative time relative to a static system (ALPS plate). The operative times, in minutes, for 14 dynamic replications of the operation and 6 static replications were obtained and are given below: Dynamic:370,360,510,445,295,315,490,345,450,505,335,280,325,500

Static :430,445,455,455,490,535

(a) Graph the data as you see fit. Why did you choose the graph(s) you did and what does it (do they) tell you? Also calculate summary statistics relevant to the research question.

(b) Perform a two population t test for means at the 5% level not assuming equal variance and justify why the assumptions of the test are reasonably met (or describe what assumptions we are assuming are met). As part of this test, specify your hypotheses, calculate a p value and make a conclusion in the context of the question. Compute the p value just using your calculator and t table. Then find an exact p value by running the same test in R.

(c) Construct and interpret an 95% confidence interval for the true difference in means (by hand and then again in R) applying the same assumptions as in part b. Describe how this confidence interval relates to your findings in part b.

(d) Suppose after the data is collected it comes to your attention that Surgeon D performed all of the Dynamic system operations and Surgeon S performed all of the Static operations. What, if any, concerns do you have about this method of data collection?

Answer 1

x <- c(370,360,510,445,295,315,490,345,450,505,335,280,325,500)
y <- c(430,445,455,455,490,535)

boxplot(x,y)
summary(x)
summary(y)

the graph chosen is boxplot

it shows the distribution of data on same plot

x- Dynamic

y- Static

summary(x)
   Min. 1st Qu. Median    Mean 3rd Qu.    Max.
280.0   327.5   365.0   394.6   480.0   510.0
> summary(y)
   Min. 1st Qu. Median    Mean 3rd Qu.    Max.
430.0   447.5   455.0   468.3   481.2   535.0

we can see that static has higher mean than that

B)

t.test(x,y,var.equal=FALSE)

        Welch Two Sample t-test

data: x and y
t = -2.6804, df = 17.832, p-value = 0.01536
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-131.48826 -15.89269
sample estimates:
mean of x mean of y
394.6429 468.3333

p-value = 0.01536 < 0.05

hence we reject the null hypothesis

C)

95 percent confidence interval:
-131.48826 -15.89269

C