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

The result obtained is: 6.0077, 215.5437

Step-by-step explanation:

the previous questionnaire is incomplete, attached and answered as requested

with the data of the exercise we obtain this information

nx-bar s

Research17593.5882160.7008

Primary care16482.8125181.2292

the exercise invites us to find the differences between the measures of a random sample of enrollments in medical schools where some specialize in research and others in primary care, with this we know that the variations in the population are different

Point estimate = 593.5882-482.8125 = 110.7757

Given the:

df = 16-1 = 15

we found that:

t * = tinv (0.1,15) = 1,753

We observe that we obtain a margin of error like this:

Margin of error = 1,753 * SQRT ((160.7008 ^ 2/17) + (181.2292 ^ 2/16)) = 104.7680

taking into account a 90% confidence interval

= 110.7757 +/- 104.768

The result obtained is: 6.0077, 215.5437

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)