Question

# # of Patients Year Month 2015 2016 1 4825 5262 2 4752 5435 3 4948 5481...

 # 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

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 H0 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.

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