Welch Two Sample t-test data: Yellowness by Feather t = 1.5832, df = 37.986, p-value = 0.1217 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -1.299296 10.623962 sample estimates: mean in group new mean in group old 143.3000 147.9623 a) From the results above, what is the null hypothesis and what statistical conclusion do you make concerning it?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 4BGP
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1. Birds can regrow feathers if they are lost. A researcher was interested in whether newly regrown feathers in
the tails of northern flickers differed in colour (yellowness in particular) compared to older feathers. To test this,
they measured the yellowness of one new and one randomly selected old feather on each of 20 birds. A plot of
their data and the results of their analysis are shown below.
Yellowness
160
150
140
130
new
X
Feather
old
Welch Two Sample t-test
=
data: Yellowness by Feather
t = 1.5832, df 37.986, p-value
165
sample estimates:
mean in group new mean in group old
147.9623
143.3000
160
New
155
150
145
140
135
= 0.1217
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.299296 10.623962
T
T
T
T
T
130 135 140 145 150 155 160
Old
a) From the results above, what is the null hypothesis and what statistical conclusion do you make concerning it?
Transcribed Image Text:1. Birds can regrow feathers if they are lost. A researcher was interested in whether newly regrown feathers in the tails of northern flickers differed in colour (yellowness in particular) compared to older feathers. To test this, they measured the yellowness of one new and one randomly selected old feather on each of 20 birds. A plot of their data and the results of their analysis are shown below. Yellowness 160 150 140 130 new X Feather old Welch Two Sample t-test = data: Yellowness by Feather t = 1.5832, df 37.986, p-value 165 sample estimates: mean in group new mean in group old 147.9623 143.3000 160 New 155 150 145 140 135 = 0.1217 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -1.299296 10.623962 T T T T T 130 135 140 145 150 155 160 Old a) From the results above, what is the null hypothesis and what statistical conclusion do you make concerning it?
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