Question

An Irate patient complained that the cost of a doctor's visit was too high. She randomly surveyed n=20 other patients and found that the mean amount of money they spent on each doctor's visit was x=$44.80. The standard deviation of the sample was s=$3.53. Find the 95% confidence interval of the population mean. Assume the variable is normally distributed.

Answer 1

zscore = (sample mean - claim)/ modified standard deviation i believe

Answer 2

claim is vague but it was just cause we went over hypot testing :)

Answer 3

2.575; 1.96; and 1.645 i recall; 95% has a zscore of 1.96

Answer 4

E = z(mu)/(sd/sqrt(n))

Answer 5

or is it mu^2?

Answer 6

mu +- E is the confidence interval

Answer 7

(44.8-mu)/(3.53/sqrt(20))) = z-score.. now rearranging you get mu = 44.8\[\pm\] Z-score(3.53)/sqrt(20)...i believe the z-score is 1.96. so the confidence interval is [43.25 ; 46.35]

Answer 8

just to recall:

n = (z o~/E)^2

sqrt(n) = z o~/E

E = z o~/sqrt(n) ; yeah, that got it lol

Answer 9

The confidence interval is 43.25

Answer 10

1.96 ; z
3.53 ; o~
-------
588
980
588
------
7.9188

hmmm, the thing is are we sure we use the zscore and not the student t?

Answer 11

im pretty sure the answer i gave is right. confidence interval [43.25 ; 46.35]

Answer 12

im sure it is too; just curious if it fits the z or t tho .. since n<30

Answer 13

if sd known use z ; got it :)

Answer 14

u cud be right,the t is used if pop SD is unknown, and we are given sample SD and not pop SD. but the fact that n>30 only comes into play if u wish to approximate one distribution using another

Answer 15

t = (sample mean-mu)/(sample SD/sqrt(n))

Answer 16

the t-interval gets me:

(43.15, 46.45) on my ti83

Answer 17

and (43.25,46.35) on the z-interval