A group of 141 terminally sick cancer patients are used in an experiment to meas

Question

A group of 141 terminally sick cancer patients are used in an experiment to measure the survival times in months of two drugs. Drug 1 is given to 79 patients selected at random, and the remaining 62 are given Drug 2. The means and standard deviations of the responses are (a) Determine a 99% confidence Interval of the mean difference of drug effects. (b) Test if the two drugs are equivalent or not (i) Specify H_0 and H_1 (b) Mate the test statistic and the rejection region with alpha = .05. (iii) Use critical value method to draw your conclusions. (iv) Use the p-value method to draw conclusions

Explanation / Answer

Mean days of survival for drug 1 x1- bar = 109

Std. Deviation s1= 21.2

N1= 79

Mean days of survival for drug 2 x2- bar = 128

Std. Deviation s1= 27.4

N2= 62

(a) 99% confidence interval for mean difference for drug effects.

99% confidence interval = (X2-bar - X1- bar) +- t/2,139 (SE0)

Here (X2-bar - X1- bar) = 128 -109 = 19

t/2,139 = t0.005,139 = +-2.611

We can z - value also because dF is more than 40

SEx1-x2 = sqrt [ s21 / n1 + s22 / n2 ] (standard deviation of either population is unknown and the sample sizes (n1 and n2) )

= sqrt [ (21.2)2 /79 + (27.4)2 /62] = 4.2188

99% confidence interval = (X2-bar - X1- bar) +- t/2,139 (SE0)

= 19 +- 2.611 * 4.2188

= (7.9847, 30.0152)

(b) Ho : There is no difference in mean survivial times for both drugs.I= II

H1: There is significnat difference in mean survaival times for both drugs. (I II)

(ii) Test Statistic

t = (X2-bar - X1- bar)/ SE0 = 19/ 4.2188 = 4.5036

and tcritical= 1.9771

so t > tcritical, we can reject the null hypothesis.

The P-Value is .000014.

The result is significant at p < .05.

So, by critical value method plus p - value method we can conclude that there is significant difference in mean survival times with drug 1 and drug 2.

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