Matched Pairs Hypothesis Test: Golf Clubs and Vitamin C in Cooked Blends - Prof. Patrick S, Study notes of Data Analysis & Statistical Methods

Examples of matched pairs hypothesis tests using t-distributions for the comparison of titleist and calloway golf clubs and vitamin c content in wheat soy blend before and after cooking. The tests determine if there is a significant difference between the means of the two populations.

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Uploaded on 08/03/2009

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Matched Pairs (Dependent Samples)
t-distibutions: confidence intervals and hypothesis tests
Matched Pairs are when two samples are taken from two populations such that the samples are
dependent on each other and can be matched up. This is not a two-sample distribution, rather, we treat
the differences of the two samples as one sample. We therefore apply t-procedures to the one
population of differences.
1. Imagine that you want to compare two sets of golf club brands: Titleist and Calloway. Suppose
that we want to show that Titleist hits farther than Calloway. We do this by comparing 10 random
pairs of golf clubs where each pair consists of one Titleist club and one Calloway club of the same
type. Each pair is hit by a different randomly chosen professional golfer (this makes the pairs
dependent since one pair can be matched up through the one golfer hitting those two clubs).
Measurements in the number of yards hit are recorded below:
Golfer Titleist Calloway Difference
1 188 194 -6
2 201 197 4
3 195 195 0
4 195 190 5
5 202 195 7
6 175 178 -3
7 182 172 10
8 188 184 4
9 193 201 -8
10 194 192 2
Is this sample of differences sufficient evidence to claim that Titleist hits farther than Calloway?
Choose 05.0=
α
.
ANSWER:
First note that the first population consists of the number of yards a Titleist club is hit, and the
second population consists of the number of yards a Calloway club is hit. Therefore, denote
T
μ
= the average number of yards a Titleist club hits
C
μ
=the average number of yards a Calloway club hits
We will assume CT
H
μ
μ
=
:
0
and we want to show CTa
H
μ
μ
>:
It is important to note that we always subtracted Calloway from Titleist in our sample. If we subtract
off C
μ
then it follows that
We will assume 0:
0
=
CT
H
μ
μ
and we want to show 0: >
CTa
H
μ
μ
Define the average of the population of differences as CTd
μ
μ
μ
=
. Then
We will assume 0:
0
=
d
H
μ
and we want to show 0: >
da
H
μ
1
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Matched Pairs (Dependent Samples) t-distibutions: confidence intervals and hypothesis tests

Matched Pairs are when two samples are taken from two populations such that the samples are dependent on each other and can be matched up. This is not a two-sample distribution, rather, we treat the differences of the two samples as one sample. We therefore apply t-procedures to the one population of differences.

  1. Imagine that you want to compare two sets of golf club brands: Titleist and Calloway. Suppose that we want to show that Titleist hits farther than Calloway. We do this by comparing 10 random pairs of golf clubs where each pair consists of one Titleist club and one Calloway club of the same type. Each pair is hit by a different randomly chosen professional golfer (this makes the pairs dependent since one pair can be matched up through the one golfer hitting those two clubs). Measurements in the number of yards hit are recorded below:

Golfer Titleist Calloway Difference 1 188 194 - 2 201 197 4 3 195 195 0 4 195 190 5 5 202 195 7 6 175 178 - 7 182 172 10 8 188 184 4 9 193 201 - 10 194 192 2

Is this sample of differences sufficient evidence to claim that Titleist hits farther than Calloway?

Choose α = 0. 05.

ANSWER:

First note that the first population consists of the number of yards a Titleist club is hit, and the second population consists of the number of yards a Calloway club is hit. Therefore, denote

μ T = the average number of yards a Titleist club hits μ (^) C =the average number of yards a Calloway club hits

We will assume H 0 : μ (^) T = μ C and we want to show Ha : μ (^) T > μ C

It is important to note that we always subtracted Calloway from Titleist in our sample. If we subtract

off μ (^) C then it follows that

We will assume H 0 : μ T − μ C = 0 and we want to show Ha : μ T − μ C > 0

Define the average of the population of differences as μ (^) d = μ T − μ C. Then

We will assume H 0 : μ d = 0 and we want to show Ha : μ d > 0

Steps conducted for the golf club Hypothesis Test:

1. Let μ d = μ T − μ C.

0 :^0

a d

d

H

H

  1. Statistics taken from the difference column ONLY!

x = 1. 5

Sx ≈ 5. 74

n = 10

10

  1. 74

= −^0 = − ≈

n

S (^) x

x

t

P-value = P (^ t^ ≥^0.^8264 )= tcdf(0.8264, 1E99, 9) ≈ 0.

Also, STAT →→TESTS 2 (T-Test) yields same results.

5. P-value> α

do NOT Reject H 0

  1. Our sample is NOT significant evidence that the average number of yards hit by a Titleist club is greater than the average number of yards hit by a Calloway club.
  2. Some researchers studying vitamin C in a corn soy blend (CSB) were also interested in a similar commodity called wheat soy blend (WSB). Both of these commodities are mixed with other ingredients and cooked. Loss of vitamin C as a result of this process was another concern of the researchers. One preparation used in Haiti called gruel (or “bouillie” in Creole) can be made from WSB, salt, sugar, milk, banana, and other optional items to improve the taste. Samples of gruel prepared in Haitian households were collected. The vitamin C content (in milligrams per 100 grams of blend, dry basis) was measured before and after cooking. Here are the results:

Sample 1 2 3 4 5 Before 73 79 86 88 78 After 20 27 29 36 17

Carry out a significance test for these data at 5% to see if vitamin C is lost after cooking.