In a random sample of 70 people, it was found that 44 of them were fans of the New York Yankees. What is the margin of error for the true proportion of all individuals who are fans of the New York Yankees? A. 0.0088 B. 0.058 C. 0.063 D. 0.116 E. 0.173
@jim_thompson5910
any ideas?
does it mention the confidence level at all?
that seems to be missing
no lol, thats exactly what it says
1 sec
ok
thats all it shows
ok let me think
alright
maybe there's other context somewhere? the reason why the confidence level is important is because the larger the confidence level, the larger the margin of error is
theres nothing else, maybe the confidence level is one of the choices, idk
for instance, if the confidence level was 95%, then the margin of error would be ... E = C*sqrt( (p*(1-p))/n ) E = 1.96*sqrt( (44/70*(1-44/70))/70 ) E = 0.11319363939728 E = 0.11319
but the confidence level is unknown
should i just guess..
it could be they're ignoring the critical value (based on the confidence level) this is obviously wrong, but oh well
sqrt( (p*(1-p))/n ) sqrt( (44/70*(1-44/70))/70 ) 0.05775185683534 0.058 that's one possibility
its not like it even really matters, i just wanna get it right to have it right, ill just go with that.
The mean biology score of all undergraduate students at a college in a given year is 67.8, with a standard deviation of 11.5. What is the standard error of the mean for a sample of 70 students?
is there a formula for that
use SE = sigma/sqrt(n)
sigma?
sigma is the standard deviation
oh lol sorry i suck at stat
\[\large \text{sigma} = \sigma\]
its fine, that's why you practice
true lol, so much formulas to remember though..
11.5/67.8(70) ?
0.116
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