4.) A medical researcher wants to determine whether a drug changes the body\'s t

Question

4.) A medical researcher wants to determine whether a drug changes the body's temperature Seven test subjects are randomly selec and the body temperature (in after 20 The subjects given and, minutes, the body temperature of each is measured again. The result are listed below At a 0.05, is there enough evidence to conclude that the drug changes the body's temperature? Assume the body temperature is normally distributed. Subject 2 5 95.5 98.1 99.4 98.9 100.2 97.9 101.8 Initial Temperature 99.2 98.4 98.2 99.0 97.8 99.7 98.6 Second Temperature Directions: For the problems above, answer the following questions. a. Identify the claim and state Ho and Ha. b. Identify the level of significance a and the degrees of freedom. c. Find the critical value and identify the rejected region. d. Calculate d and Sa e. Find the standardized test statistic t. Sketch the graph. f Decide whether to reject the null hypothesis. g. Interpret the decision in the context of the original problem.

Explanation / Answer

Given that,
population mean(u)=98.1
sample mean, x =99.4
standard deviation, s =98.9
number (n)=100.2
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 0.56
We have d = 0.56
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 7.29-(3.9^2/7 ] / 6 = 0.92
to = d/ (S/n) = 1.6
Critical Value
The Value of |t | with n-1 = 6 d.f is 2.447
We got |t o| = 1.6 & |t | =2.447
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.6043 ) = 0.1598
hence value of p0.05 < 0.1598,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 1.6
critical value: reject Ho, if to < -2.447 OR if to > 2.447
decision: Do not Reject Ho
p-value: 0.1598

X Y X-Y (X-Y)^2 101.8 99.2 2.6 6.76 98.5 98.4 0.1 0.01 98.1 98.2 -0.1 0.01 99.4 99 0.4 0.16 98.9 98.6 0.3 0.09 100.2 99.7 0.5 0.25 97.9 97.8 0.1 0.01 3.9 7.29

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