Respuesta :
First of all, I must confess that I've never studied or used this stuff, so the following in-depth analysis is brought to you direct from the seat of my pants:
-- If the portion of voters is in the interval (0.39, 0.47),
then it's 0.43 ± 0.04 .
It looks like the survey predicts 43% with a margin of error of ±4% .
-- 1500 consumers were surveyed.
46% of them amounts to 690 people.
52% of them amounts to 780 people.
The interval for the number of consumers
who are likely to prefer Drink-A is (690, 780) .
Note:
I have no confidence in my answers, and you should take the hint.
Past results are not a guarantee of future performance.
Your results depend on how you drive.
Do not send money.
This offer is void where fraud is prohibited by law.
Not a member of FDIC.
-- If the portion of voters is in the interval (0.39, 0.47),
then it's 0.43 ± 0.04 .
It looks like the survey predicts 43% with a margin of error of ±4% .
-- 1500 consumers were surveyed.
46% of them amounts to 690 people.
52% of them amounts to 780 people.
The interval for the number of consumers
who are likely to prefer Drink-A is (690, 780) .
Note:
I have no confidence in my answers, and you should take the hint.
Past results are not a guarantee of future performance.
Your results depend on how you drive.
Do not send money.
This offer is void where fraud is prohibited by law.
Not a member of FDIC.
Answer:
a) The error margins for p = 0.39 and p = 0.47 respectively are [tex]\frac{0.956}{\sqrt{n} }[/tex] and [tex]\frac{0.978}{\sqrt{n} }[/tex]
b1) number of consumers that prefer drink A = 690
b2) number of consumers that prefer drink A = 780
Step-by-step explanation:
Formula for error margin:
[tex]Error margin = z\sqrt{\frac{p(1-p)}{n} }[/tex]
Let us assume a 96% confidence level. For 96% confidence level, z = 1.96
When true estimate, p = 0.39, the error margin is:
[tex]Error margin = 1.96\sqrt{\frac{0.39(1-0.39)}{n} }\\Error margin = 1.96\sqrt{\frac{0.2379}{n} }\\Error margin = 1.96*\frac{0.4878}{\sqrt{n} } \\Error margin = \frac{0.956}{\sqrt{n} }[/tex]
When the true estimate, p = 0.47, the error margin is:
[tex]Error margin = 1.96\sqrt{\frac{0.47(1-0.47)}{n} }\\Error margin = 1.96\sqrt{\frac{0.2491}{n} }\\Error margin = 1.96*\frac{0.499}{\sqrt{n} } \\Error margin = \frac{0.978}{\sqrt{n} }[/tex]
The error margins for p = 0.39 and p = 0.47 respectively are [tex]\frac{0.956}{\sqrt{n} }[/tex] and [tex]\frac{0.978}{\sqrt{n} }[/tex]
b) Sample size = 1500
46% consumers prefer drink A
number of consumers that prefer drink A = 46% of 1500
number of consumers that prefer drink A = [tex]\frac{46}{100} * 1500[/tex]
number of consumers that prefer drink A = 690
52% of consumers prefer drink B
number of consumers that prefer drink A = 52% of 1500
number of consumers that prefer drink A = [tex]\frac{52}{100} * 1500[/tex]
number of consumers that prefer drink A = 780
