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working for two grouped means t-test
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##Q4 | |
## Refer to the setting of the previous question. | |
## To further test the system, | |
## administrators selected 20 nights and | |
## randomly assigned the new triage system to be used on | |
## 10 nights and the standard system on the remaining 10 nights. | |
## They calculated the nightly median waiting time (MWT) to see | |
## a physician. The average MWT for the new system was | |
## 3 hours with a variance of 0.60 while | |
## the average MWT for the | |
## old system was 5 hours with a variance of 0.68. | |
## Consider the 95% confidence interval estimate for the | |
## differences of the mean MWT associated with the new system. | |
## Assume a constant variance. What is the interval? | |
## Subtract in this order (New System - Old System). | |
mean_new_system = 3 | |
sample_size_new_system = 10 | |
variance_new_system = 0.6 | |
degree_new_system = sample_size_new_system - 1 | |
mean_old_system = 5 | |
sample_size_old_system = 10 | |
variance_old_system = 0.68 | |
degree_old_system = sample_size_old_system - 1 | |
sample_sizes_of_both_minus_2 = sample_size_new_system + sample_size_old_system -2 | |
## calculate the pooled variance | |
sp <- sqrt((degree_new_system * variance_new_system + degree_old_system * variance_old_system)/(sample_size_new_system + sample_size_old_system - 2)) | |
mean_new_system - mean_old_system + c(-1, 1) * qt(0.975, sample_sizes_of_both_minus_2) * sp * (1/sample_size_new_system + 1/sample_size_old_system)^0.5 | |
## answer i get is | |
## [1] -2.602492 -1.397508 |
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