Question

# of
Patients |
||

Year |
||

Month |
2015 |
2016 |

1 | 4825 | 5262 |

2 | 4752 | 5435 |

3 | 4948 | 5481 |

4 | 5041 | 6081 |

5 | 5300 | 5936 |

6 | 5232 | 5627 |

7 | 5151 | 5856 |

8 | 5585 | 5763 |

9 | 5520 | 5924 |

10 | 5136 | 5787 |

11 | 5077 | 5474 |

12 | 5300 | 5699 |

d. You set your level of significance at alpha = 0.05 and determine your t-critical value of 2.074 since you are decided to conduct a t-test. | ||||||||||||

What is your decision rule (state when would you reject the null hypothesis and when you would fail to reject the null hypothesis) | ||||||||||||

e. Using Data Analysis in Excel, you obtain the following table. Make a conclusion. Your conclusion should have two parts. | ||||||||||||

1) Do you reject or fail to reject the null hypothesis based on your decision rule? | ||||||||||||

2) Answer the question (Is there a difference in the average number of patients being seen in the emergency room between 2015 and 2016?) based on your decision. | ||||||||||||

t-Test: Two-Sample Assuming Unequal Variances | ||||||||||||

2016 |
2015 |
|||||||||||

Mean | 5693.75 | 5155.583333 | ||||||||||

Variance | 59352.205 | 63610.44697 | ||||||||||

Observations | 12 | 12 | ||||||||||

Hypothesized Mean Difference | 0 | |||||||||||

df | 22 | |||||||||||

t Stat | 5.3164387 | |||||||||||

p-value--> | P(T<=t) two-tail | 2.457E-05 | ||||||||||

t Critical two-tail | 2.0738731 |

Answer #1

To test whether there is a difference in the average number of patients being seen in the emergency room between 2015 and 2016,

i.e.to test against

Here,

The value of the test statistic

and P-value = 0

Since P-value < 0.05, so we reject H_{0} at 5% level
of significance and we can conclude that there is significant
difference in the average number of patients being seen in the
emergency room between 2015 and 2016.

# of
Patients
Year
Month
2015
2016
1
4825
5262
2
4752
5435
3
4948
5481
4
5041
6081
5
5300
5936
6
5232
5627
7
5151
5856
8
5585
5763
9
5520
5924
10
5136
5787
11
5077
5474
12
5300
5699
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