Question

om medical schools that specialize in research and from oted for primary care is listed. Find the 90% confidence interval for the 52. A random sample of enrollments from medical schools those that are noted for primary care is lis difference in the means. Research Primary Care 474 577 605 663 783 605 783 567 670 414 546 474 813 443 565 696 442 587 692 694 277 419 662 555 884 427 371 293 527 728 107 277 320

Answer 1

From the above data, we get the following results which we use to construct a 90% confidence interval for difference of means , μ_1​−μ₂ .

Source Console Research<-c(474,577,605,663,783,567,670,414,813,443,565,696,692,694,277,419,884) mul<-mean (Research) sl<-sd(Research) n1<-length (Res earch) print (mul) [1] 602.1176 print(s1) [1] 160.0589 print (n1) [1] 17 Primary_Care<-c(783,605,427,728,546, 474,371,107,442 ,587,293,277,662,555,527,320) mu2<-mean (Primary_Care) s2<-sd(Primary_Care) n2<-length(Primary_Care) print (mu2) [1] 481.5 print (s2) [1] 179.3957 > print (n2) [1] 16

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df=n1​+n2​−2=17+16−2=31.

The critical value for α=0.1 and df = 31 is t_{1−α/2;n−1}​=1.696.

The pooled standard deviation is computed as follows:

(n1 1)s(n2- 1)s Sp 2 n1n2 (17 1) x 160.0592 (16 - 1) x 179.3962 1716 2 169.691

The standard error is computed as follows:

1 1 Sp se n2 1 1 169.691 x 17 16 59.106 +

Now, finally the confidence interval is computed as follows :

(X1 - X2 te x se, Xi - X2te x se) 481.51.696 x 59.106) |(602.118 48 481.5 1.696 x 59.106, 602.118 (20.402, 220.833)