Question

A nurse at a health clinic wanted to see if its ear thermometers and oral thermometers measure the same body temperature. 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 α = 0.01?

ear	oral
97.4 97.9 96.3 97.4 98.1 97.7	97.8 98.6 98.9 97.6 99.1 98.2

a) What is the appropriate test statistic?
---Select---naz-testOne-Sample t-testIndependent-Samples t-testRelated-Samples t-test

b)
Condition 1:
---Select---ear thermometerhealth clinicoral thermometerbody temperaturestaff members
Condition 2:
---Select---ear thermometerhealth clinicoral thermometerbody temperaturestaff members

c) Input the appropriate value(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 H0Fail 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---natrivial effectsmall effectmedium effectlarge effect
r² =_______- ; ---Select---natrivial effectsmall effectmedium effectlarge 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

The given data is about the temperatures measured from the ear and oral from the patients by a nurse.

Hence, the data here in each category is drawn from related samples as a result, we should employ related t-test.

The given data is:

ear	oral
97.4	97.8
97.9	98.6
96.3	98.9
97.4	97.6
98.1	99.1
97.7	98.2

We then compute the difference in the temperatures.

ear	oral	Difference (d)
97.4	97.8	0.4
97.9	98.6	0.7
96.3	98.9	2.6
97.4	97.6	0.2
98.1	99.1	1
97.7	98.2	0.5

Next, we compute d2.

ear	oral	Difference (d)	d^2
97.4	97.8	0.4	0.16
97.9	98.6	0.7	0.49
96.3	98.9	2.6	6.76
97.4	97.6	0.2	0.04
98.1	99.1	1	1
97.7	98.2	0.5	0.25

Next, we compute the sums of each column. Doing so, we get:

	ear	oral	Difference (d)	d^2
	97.4	97.8	0.4	0.16
	97.9	98.6	0.7	0.49
	96.3	98.9	2.6	6.76
	97.4	97.6	0.2	0.04
	98.1	99.1	1	1
	97.7	98.2	0.5	0.25
SUM	584.8	590.2	5.4	8.7

The formula for paired t-statistic is:

where n is the number of data points. Here, n = 6

Substituting the values in the formula, we get:

The degrees of freedom are df = n - 1 = 6 - 1 = 5

The level of significance given is alpha = 0.01

Using a t-table, we get a critical value of 4.032

Since, t = 2.5156<4.032, we cannot reject the null hypothesis.

Hence, we conclude that the difference is not significant.

Therefore, there was no significant difference between the temperatures measured from the ear and the oral cavity.