statistics help The following data, recorded in days, represent the length of ti

Question

statistics help

The following data, recorded in days, represent the length of time to recovery for patients randomly treated with one of two medications to clear up severe bladder infections: a. Find a 95% confidence interval for the difference in the mean recovery time (mu_2 - mu_1) for the two medications, assuming normal populations. b. The producer of Medication 2 claims that the patient recovery time of their medication is faster than Medication 1. A FDA statistician wants to test this claim. What are the null and alternative hypotheses? With alpha = 5%, what is the conclusion for this test?

Explanation / Answer

(a)

n1 = 20

n2 = 15

x1-bar = 16

x2-bar = 20

s1 = 1.8

s2 = 1.5

% = 95

Degrees of freedom = n1 + n2 - 2 = 20 + 15 -2 = 33

Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((20 - 1) * 1.8^2 + ( 15 - 1) * 1.5^2)/(20 + 15 -2)) = 1.679285562

SE = Pooled s * ((1/n1) + (1/n2)) = 1.67928556237467 * ((1/20) + (1/15)) = 0.573585216

t- score = 2.034515287

Width of the confidence interval = t * SE = 2.03451528722141 * 0.573585215987999 = 1.16696789

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -4 - 1.16696789045178 = -5.16696789

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -4 + 1.16696789045178 = -2.83303211

The 95% confidence interval is [-5.17, -2.83]

(b) Ho: 2 = 1 versus Ha: 2 < 1

Data:      

n1 = 20     

n2 = 15     

x1-bar = 16     

x2-bar = 20     

s1 = 1.8     

s2 = 1.5     

Hypotheses:      

Ho: 1 2      

Ha: 1 < 2      

Decision Rule:      

= 0.05     

Degrees of freedom =   20 + 15 - 2 = 33   

Critical t- score =   -1.692360258    

Reject Ho if t <   -1.692360258    

Test Statistic:      

Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =      (((20 - 1) * 1.8^2 + (15 - 1) * 1.5^2) / (20 + 15 - 2)) = 1.679285562

SE = s * {(1 /n1) + (1 /n2)} = 1.67928556237467 * ((1/20) + (1/15)) = 0.573585216    

t = (x1-bar -x2-bar)/SE = (16 - 20)/0.573585215987999 = -6.973680438    

p- value = 2.83484E-08     

Decision (in terms of the hypotheses):      

Since -6.973680438 < -1.692360258 we reject Ho and accept Ha

Conclusion (in terms of the problem):      

The claim is true. Medication 2 acts faster than Medication 1.

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