So a simpler way than all the math being done is understanding what CI means.
CI - range of values w/in which the true mean of the population is expected fall
So a CI of 95% will be more precise and have a narrow range compared to a CI of 99% will be less precise because its including more values in and result in a wider range.
So if CI of 95% is 110 to 116 then a CI of 99% has to be a range that is wider... 108 to 118
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hTsi eitpant sah an veeraag fo 1,13 nda a %95 nodenccife ta 1011-61 amens ahtt the DS si 15. . oS one niddaatilo SD dlowu evig us a gnare of 5.,10.5181-7 rnuoded ot -.101188
VMN ogt it.
sJut YtF:eIh CI swa ttaeds ot be mrfo 116-101 iwth 9%5 nda anem of 3.S o11, no erihte teher era owt DS on ethier issed fo 311 et(h anme) taht gevi eht %.59
1=11-631 3 hwinti S2D obeav hte neam -011 11=3 3 nwihti SD2 wlboe hte aemn
3 ddvdiie by the 2 DS = 1.5 rep .SD
to get frmo 95% to %99 uoy ahve ot rcionepaotr neo oerm DS 3( DS) no rteeih idess fo teh mnea 1)31(
Teof;eehrr at %99 CI 10=15-1. 015.8 IC 11651+=. .7511
udRno ethse up dna yuo egt 811108-
For all the ~math~ people out here:
CI = mean + Z(SE)
For a 95% CI, Z = 1.96. For a 99% CI, Z = 2.58.
The CI is +3/-3 from the mean, so 3 = 1.96(SE) with the 95% CI. Solve for SE (which doesn't change if you change the CI), which comes out to about 1.53.
Now switch up the Z value for the 99% CI, with the SE you just calculated. CI = 2.58 * 1.53 = 3.95. Add this to both sides of the mean (113), and you get the answer!
I tried to calculate it more precise, and messed up the answer...
Here is why:
1 SD = 1.5 mmHg → 2.5 SD = 3.75 mmHG
This results in a 99% CI of 109.25 (113-3.75) to 116.75 (113+3.75)
Closer to answer C than B.