Question

very confused, need assistants!!!
In a random sample of 75 individuals, it was found that 52 of them prefer coffee to tea. What is the margin of error for the true proportion of all individuals who prefer coffee?

Answer 1

anything will help, need help guiding along

Answer 2

0.006

0.053

0.084

0.106

Answer 3

i think it's b?

Answer 4

@SithsAndGiggles how do i calculate what i need help ):

Answer 5

how did u get that @AriPotta

Answer 6

\[p=\frac{52}{75}=.693\]
\[1-p=.307\]
\[\text{standard error}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{.693(.307)}{75}}\approx.053\]

Answer 7

like that ^

Answer 8

hmm it said it was wrong but its happen before that i was right and still marked me wrong
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Answer 9

σn√, where σ is the standard deviation of the population and n is the sample size;

Answer 10

When you're dealing with proportions, the standard error is \(\sqrt{\dfrac{p(1-p)}{n}}\).

Answer 11

i took probability and statistics this last year, here are my notes on proportions. hopefully they help a bit

Answer 12

The 'margin of error' is the product of the z-value and the 'standard error'. To assign a z-value it is necessary to specify a percentage confidence level; for example z is 1.960 when finding a 95% confidence level.
@jcr268 Does the full question state a percentage confidence level?

Answer 13

@kropot72 True, I assumed @jcr268 meant standard error because the CL was left out.

Answer 14

@SithsAndGiggles I see your reasoning :)

Answer 15

It's D