Statistics - constructing and interpreting confidence intervals Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Statistics Confidence IntervalsDegrees of freedom: when to use infinity?Statistics and confidence - intervals95% Confidence IntervalsStatistics - Confidence Intervals - $P(X>1)$Interpreting confidence intervalsWhat does glassware accuracy and precision (tolerance) mean statistically?Why is this the wrong way to interpret a confidence interval?Constructing joint confidence intervals/multiple confidence intervalsStatistics and Confidence Intervals
Is there hard evidence that the grant peer review system performs significantly better than random?
How can I set the aperture on my DSLR when it's attached to a telescope instead of a lens?
What are the discoveries that have been possible with the rejection of positivism?
Deconstruction is ambiguous
How much damage would a cupful of neutron star matter do to the Earth?
What would you call this weird metallic apparatus that allows you to lift people?
What does it mean that physics no longer uses mechanical models to describe phenomena?
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
What makes a man succeed?
Drawing spherical mirrors
Does "shooting for effect" have contradictory meanings in different areas?
Do wooden building fires get hotter than 600°C?
Co-worker has annoying ringtone
Would it be easier to apply for a UK visa if there is a host family to sponsor for you in going there?
How to write capital alpha?
How often does castling occur in grandmaster games?
Does the Mueller report show a conspiracy between Russia and the Trump Campaign?
Semigroups with no morphisms between them
1-probability to calculate two events in a row
How does light 'choose' between wave and particle behaviour?
How does the math work when buying airline miles?
What is the meaning of 'breadth' in breadth first search?
The Nth Gryphon Number
Putting class ranking in CV, but against dept guidelines
Statistics - constructing and interpreting confidence intervals
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Statistics Confidence IntervalsDegrees of freedom: when to use infinity?Statistics and confidence - intervals95% Confidence IntervalsStatistics - Confidence Intervals - $P(X>1)$Interpreting confidence intervalsWhat does glassware accuracy and precision (tolerance) mean statistically?Why is this the wrong way to interpret a confidence interval?Constructing joint confidence intervals/multiple confidence intervalsStatistics and Confidence Intervals
$begingroup$
I attempted to solve the following statistics questions below and I was wondering if someone would be willing to check my answers and confirm they're correct? I was also wondering if there is a way I could expand my answer for c).
According to an article in the New York Times, America's elderly paid an average of 3757 dollars for healthcare in 2003. The figure was based on a random sample of 560 elderly Americans and the standard deviation of the sample was 870 dollars.
(a) Determine an 88% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.88)/2 = 0.06
1-0.06 = 0.94
p(z < 1.56) = 0.94
x ± z(s/√n)
3757 ± ((1.56)(870/√560))
3757 - 57.35216 = 3699.65
3757 + 57.35216 = 3814.35
Therefore the confidence interval is between 3699.65 and 3814.35.
(b) Determine a 94% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.94)/2 = 0.03
1-0.03 = 0.97
p(z < 1.89) = 0.97
x ± z(s/√n)
3757 ± ((1.89)(870/√560))
3757 - 69.48435 = 3687.52
3757 + 69.48435 = 3826.48
Therefore the confidence interval is between 3687.52 and 3826.48
(c) Interpret the confidence interval constructed in part (b)
94% of elder americans will pay between 3687.52 and 3826.48 in health care in 2003.
I feel like I'm missing something, but I don't really know how to interpret the data further than that?
statistics confidence-interval
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
$endgroup$
add a comment |
$begingroup$
I attempted to solve the following statistics questions below and I was wondering if someone would be willing to check my answers and confirm they're correct? I was also wondering if there is a way I could expand my answer for c).
According to an article in the New York Times, America's elderly paid an average of 3757 dollars for healthcare in 2003. The figure was based on a random sample of 560 elderly Americans and the standard deviation of the sample was 870 dollars.
(a) Determine an 88% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.88)/2 = 0.06
1-0.06 = 0.94
p(z < 1.56) = 0.94
x ± z(s/√n)
3757 ± ((1.56)(870/√560))
3757 - 57.35216 = 3699.65
3757 + 57.35216 = 3814.35
Therefore the confidence interval is between 3699.65 and 3814.35.
(b) Determine a 94% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.94)/2 = 0.03
1-0.03 = 0.97
p(z < 1.89) = 0.97
x ± z(s/√n)
3757 ± ((1.89)(870/√560))
3757 - 69.48435 = 3687.52
3757 + 69.48435 = 3826.48
Therefore the confidence interval is between 3687.52 and 3826.48
(c) Interpret the confidence interval constructed in part (b)
94% of elder americans will pay between 3687.52 and 3826.48 in health care in 2003.
I feel like I'm missing something, but I don't really know how to interpret the data further than that?
statistics confidence-interval
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
$endgroup$
add a comment |
$begingroup$
I attempted to solve the following statistics questions below and I was wondering if someone would be willing to check my answers and confirm they're correct? I was also wondering if there is a way I could expand my answer for c).
According to an article in the New York Times, America's elderly paid an average of 3757 dollars for healthcare in 2003. The figure was based on a random sample of 560 elderly Americans and the standard deviation of the sample was 870 dollars.
(a) Determine an 88% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.88)/2 = 0.06
1-0.06 = 0.94
p(z < 1.56) = 0.94
x ± z(s/√n)
3757 ± ((1.56)(870/√560))
3757 - 57.35216 = 3699.65
3757 + 57.35216 = 3814.35
Therefore the confidence interval is between 3699.65 and 3814.35.
(b) Determine a 94% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.94)/2 = 0.03
1-0.03 = 0.97
p(z < 1.89) = 0.97
x ± z(s/√n)
3757 ± ((1.89)(870/√560))
3757 - 69.48435 = 3687.52
3757 + 69.48435 = 3826.48
Therefore the confidence interval is between 3687.52 and 3826.48
(c) Interpret the confidence interval constructed in part (b)
94% of elder americans will pay between 3687.52 and 3826.48 in health care in 2003.
I feel like I'm missing something, but I don't really know how to interpret the data further than that?
statistics confidence-interval
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
$endgroup$
I attempted to solve the following statistics questions below and I was wondering if someone would be willing to check my answers and confirm they're correct? I was also wondering if there is a way I could expand my answer for c).
According to an article in the New York Times, America's elderly paid an average of 3757 dollars for healthcare in 2003. The figure was based on a random sample of 560 elderly Americans and the standard deviation of the sample was 870 dollars.
(a) Determine an 88% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.88)/2 = 0.06
1-0.06 = 0.94
p(z < 1.56) = 0.94
x ± z(s/√n)
3757 ± ((1.56)(870/√560))
3757 - 57.35216 = 3699.65
3757 + 57.35216 = 3814.35
Therefore the confidence interval is between 3699.65 and 3814.35.
(b) Determine a 94% confidence interval for the actual average spending on health care by elderly Americans in 2003.
z=(1-0.94)/2 = 0.03
1-0.03 = 0.97
p(z < 1.89) = 0.97
x ± z(s/√n)
3757 ± ((1.89)(870/√560))
3757 - 69.48435 = 3687.52
3757 + 69.48435 = 3826.48
Therefore the confidence interval is between 3687.52 and 3826.48
(c) Interpret the confidence interval constructed in part (b)
94% of elder americans will pay between 3687.52 and 3826.48 in health care in 2003.
I feel like I'm missing something, but I don't really know how to interpret the data further than that?
statistics confidence-interval
statistics confidence-interval
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
asked Apr 2 at 0:59
GilmoreGirlingGilmoreGirling
395
395
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
$endgroup$
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171334%2fstatistics-constructing-and-interpreting-confidence-intervals%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
$endgroup$
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
add a comment |
$begingroup$
Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
$endgroup$
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
add a comment |
$begingroup$
Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
$endgroup$
Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
answered Apr 2 at 1:40
Kevin2019Kevin2019
223
223
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
add a comment |
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
$begingroup$
Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers.
$endgroup$
– Brian
Apr 2 at 1:50
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171334%2fstatistics-constructing-and-interpreting-confidence-intervals%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
