Question

Between February 1995 and December 1997, 17 consecutive patients with combined acute ACL and grade III MCL injuries were treated by the same physician at the research center. One of the variables of interest was the length of time in days between the occurrence of the injury and the first magnetic resonance imaging (MRI).

Here is the data: 14, 9, 18, 26, 12, 0, 10, 4, 8, 21, 28, 24, 24, 2, 3, 14, 9. The sample mean and sample standard deviation are 13.2941 and 8.8865, respectively.

Q1. Construct a 95% confidence interval for the population mean number of days between the occurrence of the injury and the first MRI and write a brief interpretation of your confidence interval in terms of the context.

Q2. (2 points) Under what assumptions is the confidence interval in Q1 constructed?

Q3. (3 points) Without constructing a 99% confidence interval, what can you say about the 99% confidence interval compared to the 95% one in Q1 in terms of the interval length? Write a sentence to justify your answer.

Q4. Do the data support the claim that the population mean number of days between the occurrence of the injury and the first MRI is less than 15 days at a significance level of 0.05? Answer the question by the following steps.

d) Formulate null and alternative hypotheses in terms of the population parameter as well as the context of this application;

e) State type I and type II errors in terms of the context of this application.

f) Is the test right-tailed, left-tailed or two-tailed?

g) Find the P-value of the test, interpret your P-value in the context of the application, and make your conclusion.

Answer 1

a)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   16          
't value='   tα/2=   2.120   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   8.8865/√17=   2.1553          
margin of error , E=t*SE =   2.1199   *   2.1553   =   4.5690
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    13.29   -   4.569017   =   8.7251
Interval Upper Limit = x̅ + E =    13.29   -   4.569017   =   17.8631
95%   confidence interval is (   8.73   < µ <   17.86   )

b) population from which samples are teaken is normally distributed

3)99% confidence interval is wider than that of 95% confidence interval

4)

d)

Ho :   µ =   15
Ha :   µ <   15

e) Type I error is concluding that true mean is less than 15 while infact it is 15

type II error is concluding that true mean is 15 while infact ot is false

f) left tailed

g)

sample std dev ,    s =    8.8865
Sample Size ,   n =    17
Sample Mean,    x̅ =   13.2941
      
degree of freedom=   DF=n-1=   16
      
Standard Error , SE = s/√n =   8.8865/√17=   2.1553
t-test statistic= (x̅ - µ )/SE =    (13.2941-15)/2.1553=   -0.791
      

      
p-Value   =   0.2201

Decision:   p-value>α, Do not reject null hypothesis