Question

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 < μ₁ - μ₂ < 231.2589

  

24.8659 < μ₁ - μ₂ < 196.6855

  

6.0077 < μ₁ - μ₂ < 215.5437

Answer 1

Ans:

	n	x-bar	s
Research	17	593.5882	160.7008
Primary care	16	482.8125	181.2292

Assumption:Population variances are not equal.

Point estimate=593.5882-482.8125=110.7757

df=16-1=15

t*=tinv(0.1,15)=1.753

Margin of error=1.753*SQRT((160.7008^2/17)+(181.2292^2/16))=104.7680

90% confidence interval for difference in means

=110.7757+/-104.768

=(6.0077, 215.5437)

Last option is correct.