Question

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. | |||||||||

H_{0}: |
|||||||||

H_{1}: |
|||||||||

Sample's Mean (x-bar): | |||||||||

Sample's S.D. (s): | |||||||||

Sample's Test Statistic: | |||||||||

P-value: | |||||||||

CONCLUSION: |

Answer #1

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.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 20 minutes ago

asked 20 minutes ago

asked 23 minutes ago

asked 30 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago