Question

A college admissions director wishes to estimate the mean age of all students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answers to two decimal places and use increasing order.

Age

25.8

22.2

22.5

22.8

24.6

24.0

22.6

23.6

22.8

23.1

21.5

21.4

22.5

24.5

21.5

22.5

20.5

23.0

25.1

25.2

23.8

21.8

24.1

Answer #1

Age(x) | (x-xbar)^2 | |

25.8 | 7.266541 | |

22.2 | 0.817845 | |

22.5 | 0.365236 | |

22.8 | 0.092628 | |

24.6 | 2.236975 | |

24 | 0.802193 | |

22.6 | 0.254367 | |

23.6 | 0.245671 | |

22.8 | 0.092628 | |

23.1 | 1.89E-05 | |

21.5 | 2.573932 | |

21.4 | 2.904802 | |

22.5 | 0.365236 | |

24.5 | 1.947845 | |

21.5 | 2.573932 | |

22.5 | 0.365236 | |

20.5 | 6.782628 | |

23 | 0.010888 | |

25.1 | 3.982628 | |

25.2 | 4.391758 | |

23.8 | 0.483932 | |

21.8 | 1.701323 | |

24.1 | 0.991323 | |

sum | 531.4 | 41.24957 |

mean | 23.10435 |

s=sample standard deviation

s=1.3693

#90% confidence interval for the population mean age

n=23 s=1.369

90 % confidence interval forPOPulation Mean is

ME=margin of error

t=critical value

t=

**t _{22,0.05=1.7171}**

_{}

ME=0.4901

_{}

t is obtained from t- table with corresponding degree of freedom (n-1) is22

# using Excell =TINV(0.1,22)

90 % Confidence interval for Population Mean is

(23.10-0.4901,23.10+0.4901)

(22.61,23.59)

#90% confidence interval for the population mean age. is (22.61,23.59)

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