Suppose we obtained data on vein size after application of a nitroglycerin ointment in a sample of 50 patients. the mean vein size is found to be 8.7mm with an sd of 2.1. using a t distribution table, what are the confidence limits for a 95% confidence interval? for a 99% confidence interval?

95% confidence interval results: CI between lower limit 8.103166mm and upper limit 9.2968134mmμ : Mean of . Err.DFL. LimitU. Limitμ8.70.29698485 498.10318669.296813499% confidence interval results:CI between lower limit 7.9040949mm and upper limit 9.4959051mmμ : Mean of . Err.DFL. LimitU. Limitμ8.70.29698485 497.90409499.495905

the answer

b. they are parallel and will never intersect.

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.