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 ? = .05, if there is a difference in the mean number of beats for the two groups.

Write the null and alternative hypotheses. (2 pts)

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. (2 pts)

State the test statistic and p-value of the appropriate t-test, and interpret the result. (2 pts)

The TTEST Procedure

Variable: beats

N	Mean	StdDev	StdErr	Minimum	Maximum
41	336.8	25.5234	3.9861	293.0	403.0

Mean	95% CL Mean	StdDev	95% CL Std Dev
336.8	328.8	344.9	25.5234	20.9551	32.6573

DF	tValue	Pr>|t|
40	84.50	<.0001>

Distribution of beats With 95% Confidence Interval for Mean Normal Kernel 30 20 Percent 10 0 0 95% Confidence 250 300 350 400 beats