The number of heart beats in a 5-minute period are recorded for 25 women not using a certain drug, and 16 women using this drug.

No drug:       293   297   300   302   306   311   312   314   318   320   322   323   324

                   324   326   327   331   333   335   339   342   345   345   351   356

    Drug:       330   331   335   337   339   343   346   349

                   350   358   364   369   375   387   398   403

data two;

    input drug $ beats @@;

datalines;

No 293 No 297 No 300 No 302 No 306 No 311 No 312 No 314

No 318 No 320 No 322 No 323 No 324 No 324 No 326 No 327

No 331 No 333 No 335 No 339 No 342 No 345 No 345 No 351 No 356

Yes 330 Yes 331 Yes 335 Yes 337 Yes 339 Yes 343 Yes 346 Yes 349 Yes 350 Yes 358 Yes 364 Yes 369 Yes 375 Yes 387 Yes 398 Yes 403

;

run;

We want to test, at a = .05, if there is a difference in the mean number of beats for the two groups.

a. Write the null and alternative hypotheses.

b. Analyze the data using PROC TTEST, and turn in the printout.  From this, you will first have to test if the population variances are equal.  Give the p-value of this test, then briefly state which t-test (i.e., 'Equal' or 'Unequal') is appropriate to compare the means.

c. State the test statistic and p-value of the appropriate t-test, and interpret the result.