I need the complete working solutions and accurate answers to this question. Tha
ID: 2934187 • Letter: I
Question
I need the complete working solutions and accurate answers to this question. Thank you.
(10 pts.) 2. Adding computerized medical images to a database promises to provide great resources for physicians. However, there are other methods of obtaining such information so the issue of efficiency of access needs to be investigate. The information on a an experiment in which 13 computer-proficient medical professionals were timed (in seconds) both while retrieving an image from a library of slides and while retrieving the same image from a computer database with a Web front end. Please assume that the distribution is normal. The summary data are given in the following table. Measure (in seconds) Time to retrieve data from slides Time to retrieve data from the Web Difference in Time (slide - Web) 13 36.00 1.365 32.0810.177 3.840 3.92 (1 pt.) a) Should this situation be analyzed via a 2-sample independent or paired method? Please explain your answer. If this is a pairs situation, please state the common characteristic that makes these data paired. (6 pts.) b) Is there any evidence to suggest that the mean time to retrieve the figure for the two methods is different? Use = 0.01. Remember to include the complete 4-step method in the hypothesis test. (2 pts.) c) Calculate and interpret the appropriate confidence interval or bound. (1 pt.) d) In practical terms, does the data imply that the mean time is different for the two methods? This part uses the information from parts b) and c), however, if no addition reasoning is provided, you will receive 0 pointsExplanation / Answer
a.
it is a t test for 2 sample independent. since the invidual data is not
given and the mesaurements are independent to each other it fall under
independent test
b.
Given that,
mean(x)=36
standard deviation , s.d1=11.465
number(n1)=13
y(mean)=32.08
standard deviation, s.d2 =10.177
number(n2)=13
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.055
since our test is two-tailed
reject Ho, if to < -3.055 OR if to > 3.055
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =36-32.08/sqrt((131.44623/13)+(103.57133/13))
to =0.922
| to | =0.922
critical value
the value of |t | with min (n1-1, n2-1) i.e 12 d.f is 3.055
we got |to| = 0.92195 & | t | = 3.055
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.922 ) = 0.375
hence value of p0.01 < 0.375,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.922
critical value: -3.055 , 3.055
decision: do not reject Ho
p-value: 0.375
we do not have evidence that mean time to retrieve the figure for two methods is different.
c.
given that,
mean(x)=36
standard deviation , s.d1=11.465
sample size, n1=13
y(mean)=32.08
standard deviation, s.d2 =10.177
sample size,n2 =13
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 36-32.08) ± t a/2 * sqrt((131.446/13)+(103.571/13)]
= [ (3.92) ± t a/2 * 4.252]
= [-7.479 , 15.319]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-7.479 , 15.319] contains the true population mean
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
we do not have evidence that mean time to retrieve the figure for two methods is different.
d.
From Hypothesis, we conclude do not reject Ho, which means null, Ho: u1 = u2
From CI, [-7.479 , 15.319] ., we contains the true population mean and in practical we knows that time to retrieve data is relavantly has high intensity rate as compared to the normal patients.
considering above three statement, we claim that mean is not diffrent for two methods