Question

Use the information below to answer the CANCER SURVIVAL questions that follow. Data originally from DASL http://lib.stat.cmu.edu/DASL/Datafiles/CancerSurvival.html e. Some bronchial cancer patients had the following survival times (in days) after receiving a certain treatment: 20, 37, 63, 64, 72, 81, 138, 151, 155, 166, 166, 223, 245, 246, 450, 459. (The second instance of "166" is not a mistake.) Do these numbers suggest that the average survival time, for all bronchial cancer patients who receive that treatment, is significantly different from 200 days?

Answer 1

calculate mean and standard deviation of the data:

Count, N: 16 Sum, Ex: 2736 Mean, X 171 Variance, s2: 17210.4 Steps N 1 S= (x; – T), N-1 i=1 $22 2 = E(X; - x)2 11 N-1 (20 - 171)2 + ... + (459 - 171)2 16 - 1 258156 15 = 17210.4 S = V17210.4 = 131.18841412259

assume significance level of 0.05

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 200 Ha: u 7 200 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.131. The rejection region for this two-tailed test is R= {t: |t| > 2.131} (3) Test Statistics The t-statistic is computed as follows: X-LO t= 171 - 200 131.1884/16 -0.884 8/n (4) Decision about the null hypothesis Since it is observed that t = 0.884 <te=2.131, it is then concluded that the null hypothesis is not rejected.

(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 200, at the 0.05 significance level.

we are allowed to solve one question only.