Question
In: Statistics and Probability
A 2010 poll asked people in the United States whether they were satisfied with their financial...
A
2010
poll asked people in the United States whether they were satisfied with their financial situation. A total of
338
out of
833
people said they were satisfied. The same question was asked in
2012
, and
304
out of
1156
people said they were satisfied.
Part 1 of 2
Your Answer is correct
(a) Construct a
99.8%
confidence interval for the difference between the proportions of adults who said they were satisfied in
2012
and
2010
. Let
p1
denote the proportion of adults who said they were satisfied in
2010
.Use tables to find the critical value and round the answer to at least three decimal places.
A
99.8%
confidence interval for the difference between the proportions of adults who said they were satisfied in
2012
and
2010
is
<0.077<−p1p20.209.
Part: 1 / 2
1 of 2 Parts Complete
Part 2 of 2
(b) A sociologist claims that the proportion of people who are satisfied increased from
2010
to
2012
by more than
0.22
. Does the confidence interval contradict this claim?
| Because the confidence
interval ▼(Choose one) values above
0.22 , it ▼(Choose one) the claim that the proportion of people who are satisfied increased from2010 to2012 by more than0.22 . |
Solutions
THANK
YOU
orchestra answered 1 year agoRelated Solutions
A 2010 poll asked people in the United States whether they were satisfied with their financial...
A
2010
poll asked people in the United States whether they were
satisfied with their financial situation. A total of
338
out of
833
people said they were satisfied. The same question was asked
in
2012
, and
304
out of
1156
people said they were satisfied.
Part: 0 / 2
0 of 2 Parts Complete
Part 1 of 2
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