Question
Jan Villaroel
Topic: Statistics Posted 1 year ago
The authors of a research article compared two different instruments for measuring a person's capacity to breathe out air. (This mea

The authors of a research article compared two different instruments for measuring a person's capacity to
breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments
considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people
participated in the study, and for each person, air flow was measured once using the Wright meter and once
using the mini-Wright meter. The Wright meter is thought to provide a better measure of air flow, but the
mini-Wright meter is easier to transport and to use. Use of the mini-Wright meter could be recommended
as long as there is not convincing evidence that the mean reading for the mini-Wright meter is different
from the mean reading for the Wright meter. For purposes of this exercise, you can assume that it is
reasonable to consider the 17 people who participated in this study as representative samples from their
respective populations of measurement. Data values from this paper are given in the accompanying table.
Subject Mini- Wright-Meter Wright -Meter
1 514 494
2 430 395
3 520 516
4 428 434
5 500 476
6 600 557
7 364 413
8 380 442
9 658 650
10 442 433
11 432 417
12 626 656
13 260 267
14 477 478
15 259 178
16 350 423
17 451 422


a) State the appropriate null and alternative hypotheses

b) Find the test statistic.


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Koala Tutor
Dec 14, 2023
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a) The appropriate null and alternative hypotheses for this study are:

Null hypothesis: The mean reading for the mini-Wright meter is equal to the mean reading for the Wright meter.
Alternative hypothesis: The mean reading for the mini-Wright meter is different from the mean reading for the Wright meter.

b) To find the test statistic, we will conduct a paired t-test since the same subjects were measured using both instruments. We will calculate the difference between the paired observations and perform the t-test on the differences.

The differences between the mini-Wright meter and Wright meter readings are as follows:

Difference 1: 514 - 494 = 20
Difference 2: 430 - 395 = 35
Difference 3: 520 - 516 = 4
Difference 4: 428 - 434 = -6
Difference 5: 500 - 476 = 24
Difference 6: 600 - 557 = 43
Difference 7: 364 - 413 = -49
Difference 8: 380 - 442 = -62
Difference 9: 658 - 650 = 8
Difference 10: 442 - 433 = 9
Difference 11: 432 - 417 = 15
Difference 12: 626 - 656 = -30
Difference 13: 260 - 267 = -7
Difference 14: 477 - 478 = -1
Difference 15: 259 - 178 = 81
Difference 16: 350 - 423 = -73
Difference 17: 451 - 422 = 29

Next, we calculate the mean difference and the standard deviation of the differences:

Mean difference (d̄) = (20 + 35 + 4 - 6 + 24 + 43 - 49 - 62 + 8 + 9 + 15 - 30 - 7 - 1 + 81 - 73 + 29)/17 = -1.76

Standard deviation of differences (s) = sqrt[((20 - (-1.76))^2 + (35 - (-1.76))^2 + (4 - (-1.76))^2 + (-6 - (-1.76))^2 + (24 - (-1.76))^2 + (43 - (-1.76))^2 + (-49 - (-1.76))^2 + (-62 - (-1.76))^2 + (8 - (-1.76))^2 + (9 - (-1.76))^2 + (15 - (-1.76))^2 + (-30 - (-1.76))^2 + (-7 - (-1.76))^2 + (-1 - (-1.76))^2 + (81 - (-1.76))^2 + (-73 - (-1.76))^2 + (29 - (-1.76))^2]/(17-1)]

Calculating the above expression gives us s = 36.15

The test statistic for a paired t-test is given by t = (d̄ - 0) / (s / sqrt(n)), where n is the number of paired observations.

Using the above formula, we can calculate the test statistic:

t = (-1.76 - 0) / (36.15 / sqrt(17)) ≈ -0.20

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