From the above data, we get the following results which we use to construct a 90% confidence interval for difference of means , μ1−μ2 .
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 t1−α/2;n−1=1.696.
The pooled standard deviation is computed as follows:
The standard error is computed as follows:
Now, finally the confidence interval is computed as follows :
om medical schools that specialize in research and from oted for primary care is listed. Find...
A random sample of enrollments from medical schools that specialize in research and from those that are noted for primary care is listed. Find the 90% confidence interval for the difference in the means. Research 474 571 596 658 782 477 667 414 807 437 571 688 689 692 280 414 874 Primary care 783 609 420 736 556 469 371 115 442 597 291 278 670 556 522 310 18.5799 < μ1 - μ2 < 202.9715 -9.7075...
Create a program that will use the attached input file and perform the following operations. Read the file into an appropriate JCF data structure. Look up a (list of) names and numbers matching a last name or the first letters of a last name, ignoring case. Look up a (list of) names and numbers matching a number or the first digits of a number. Add a name and number to the list. Sort the list by first name, last name...