Question

In: Statistics and Probability

It has been claimed that the mean armspan of adults in the US is 174 cm....

It has been claimed that the mean armspan of adults in the US is 174 cm. Does the data of our statistics class support or contradict this claim? Justify your answer through a formal hypothesis testing procedure with a P-value approach using a significance level of your choice. Calculation of and interpretation based on the P-value is required in this problem.

H0: µ=174
H1: µ≠174
Sample's Mean (x-bar): 171.6
Sample's S.D. (s): 14.11743539
Sample's Test Statistic: -1.476354886
P-value: 0.072045781
CONCLUSION:
Armspan
181.0
187.5
176.0
160.5
137.0
165.0
172.0
170.0
155.5
176.5
183.0
162.5
155.0
179.0
163.0
175.0
186.0
164.0
155.0
172.0
165.0
190.5
147.5
173.0
162.0
154.0
163.0
169.5
169.0
165.0
203.0
188.0
176.0
165.0
154.5
171.0
167.0
172.0
208.5
194.0
165.0
190.5
193.0
152.0
195.0
149.0
161.0
160.0
176.5
190.0
182.5
175.0
166.0
196.0
167.0
181.0
167.5
160.5
165.0
157.5
183.0
150.0
170.0
183.0
188.0
160.0
157.5
192.0
172.5
181.0
152.0
175.0
165.0
175.5
182.0

Solutions

Expert Solution

using excel>addin>phstat>one sample test

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m= 174
Level of Significance 0.05
Sample Size 75
Sample Mean 171.5933333
Sample Standard Deviation 14.11743539
Intermediate Calculations
Standard Error of the Mean 1.6301
Degrees of Freedom 74
t Test Statistic -1.4764
Two-Tail Test
Lower Critical Value -1.9925
Upper Critical Value 1.9925
p-Value 0.1441
Do not reject the null hypothesis

we have given the null and alternative hypothesis is

H0: µ=174
H1: µ≠174

test stat = -1.4764

p value is 0.1441( p value provided by you is wrong because that is for one tail but your test is two tail here )

since p value is greater than 0.05 so we do not reject Ho so we have sufficient evidance to accept the claime that the mean armspan of adults in the US is 174 cm

orchestra answered 2 years ago
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