Gender | Armspan |
Male | 174.0 |
Female | 163.0 |
Male | 200.5 |
Female | 174.0 |
Female | 162.0 |
Female | 162.5 |
Male | 193.0 |
Female | 172.5 |
Female | 163.0 |
Female | 195.5 |
Male | 170.0 |
Male | 181.0 |
Female | 168.0 |
Female | 168.0 |
Male | 185.0 |
Female | 170.0 |
Female | 168.5 |
Female | 162.0 |
Male | 169.0 |
Female | 164.5 |
Female | 165.5 |
Male | 177.5 |
Female | 174.0 |
Female | 175.0 |
Male | 174.5 |
Male | 173.0 |
Female | 160.0 |
Female | 172.0 |
Male | 180.0 |
Female | 173.0 |
Male | 178.0 |
Male | 185.5 |
Female | 154.5 |
Female | 170.0 |
Female | 170.0 |
Female | 171.0 |
Female | 157.5 |
Male | 185.0 |
Female | 161.5 |
Female | 155.0 |
Female | 160.5 |
Female | 180.5 |
Female | 162.5 |
Male | 168.0 |
Male | 169.5 |
Female | 150.0 |
Female | 152.5 |
Female | 163.0 |
Female | 158.0 |
Female | 167.5 |
Female | 152.0 |
Female | 159.5 |
Female | 160.0 |
Female | 152.5 |
Female | 157.0 |
Female | 167.5 |
Male | 177.5 |
Female | 174.0 |
Male | 192.5 |
Female | 160.0 |
Male | 193.0 |
Male | 175.0 |
Male | 184.0 |
Female | 154.0 |
Male | 194.0 |
Male | 182.5 |
Female | 142.0 |
Female | 160.0 |
Female | 157.0 |
Female | 163.0 |
Male | 188.0 |
Male | 180.0 |
Female | 150.0 |
has been claimed that the mean armspan of adults in the US is greater than 166 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 of the P-value is required on this problem. | |||||||||
H0: | |||||||||
H1: | |||||||||
Sample's Mean (x-bar): | |||||||||
Sample's S.D. (s): | |||||||||
Sample's Test Statistic: | |||||||||
P-value: | |||||||||
CONCLUSION: |
First enter Data into EXCEL
We have to find the sample mean.
Excel command is =AVERAGE(Select data)
sample mean = 169.664
Now we have to find sample standard deviation.
Excel command is =STDEV(Select data)
standard deviation = 12.314
T test
n = 73
s = 12.314
Null and alternative hypothesis is
H0 : u = 166
H1 : u > 166
Level of significance = 0.05
Here population standard deviation is not known so we use t-test statistic.
Test statistic is
Degrees of freedom = n - 1 = 73 - 1 = 72
P-value = 0.0066 ( using t table)
P-value , Reject H0
conclusion : There is sufficient evidance to conclude that the mean armspan of adults in the US is greater than 166 cm.
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