Question

A nurse at a health clinic hypothesizes that ear thermometers measure higher body temperatures than oral thermometers. The nurse selects a sample of healthy staff members and took the temperature of each with both thermometers. The temperature data are below. What can the nurse conclude with an α of 0.01?

ear	oral
97.4 97.9 96.3 97.4 98.1 97.3	97.8 98.6 98.9 97.9 99.1 98.2

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- health clinic ear thermometer body temperature oral thermometer staff members
Condition 2:
---Select--- health clinic ear thermometer body temperature oral thermometer staff members

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

The ear thermometer measured significantly higher temperatures than the oral thermometer.

The ear thermometer measured significantly lower temperatures than the oral thermometer.    

There was no significant temperature difference between the ear and oral thermometer.

Answer 1

Given that,
mean(x)=97.4
standard deviation , s.d1=0.6261
number(n1)=6
y(mean)=98.4167
standard deviation, s.d2 =0.5345
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =3.365
since our test is right-tailed
reject Ho, if to > 3.365
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =97.4-98.4167/sqrt((0.392/6)+(0.28569/6))
to =-3.0252
| to | =3.0252
critical value
the value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 3.365
we got |to| = 3.02519 & | t α | = 3.365
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > -3.0252 ) = 0.98538
hence value of p0.01 < 0.98538,here we do not reject Ho
ANSWERS
---------------
a.
Independent-Samples t-test
b.
health clinic ear thermometer body temperature oral thermometer staff members
c.
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -3.0252
critical value: 3.365
decision: do not reject Ho
p-value: 0.98538
we do not have enough evidence to support the claim that ear thermometers measure higher body temperatures than oral thermometers
d.
Effective size = modulus of (mean 1 -mean 2)/standard deviation pooled(S.Dpooled)
Effective size = modulus of (mean 1 - mean2 )/S.Dpooled
S.Dpooled = sqrt (((n1-1)*s.d1^2+(n2-1)*s.d2 ^2)/(n1+n2-2))
S.Dpooled= sqrt (((6-1)*0.6261^2 +(6-1)*0.5345^2)/(6+6-2))
S.Dpooled= 0.5821
Effective size = modulus of (97.4-98.4167)/0.5821
Effective size = modulus of (-1.7466)
Effective size = 1.7466
large effect
e.
The ear thermometer measured significantly higher temperatures than the oral thermometer.