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qveenjordan6456
26.11.2019 •
Mathematics
In a recent poll, 600 people were asked if they liked soccer, and 72% said they did. based on this, construct a 99% confidence interval for the true population proportion of people who like soccer.
as in the reading, in your calculations:
--use z = 1.645 for a 90% confidence interval
--use z = 2 for a 95% confidence interval
--use z = 2.576 for a 99% confidence interval.
give your answers as decimals, to 4 decimal places.
out of 500 people sampled, 80 had kids. based on this, construct a 90% confidence interval for the true population proportion of people with kids.
as in the reading, in your calculations:
--use z = 1.645 for a 90% confidence interval
--use z = 2 for a 95% confidence interval
--use z = 2.576 for a 99% confidence interval.
give your answers to three decimals
give your answers as decimals, to three decimal places.
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Ответ:
(0.6728, 0.7672)
(0.1330, 0.1870)
Step-by-step explanation:
Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.
Std error =![\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183](/tpl/images/0391/1640/4c5de.png)
Margin of error for 99% we would use the value![z = 2.576](/tpl/images/0391/1640/1a296.png)
Margin of error =![2.576*SE\\=2.576*0.0183\\=0.0472](/tpl/images/0391/1640/4db77.png)
Confidence interval lower bound =![0.72-0.0472=0.6728](/tpl/images/0391/1640/2851e.png)
Upper bound =![0.72+0.0472=0.7672](/tpl/images/0391/1640/d4204.png)
99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)
b) n =500
Sample proportion p=![\frac{80}{500} =0.16](/tpl/images/0391/1640/86d71.png)
Margin of error for 90% =![1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270](/tpl/images/0391/1640/f9b1a.png)
90% confidence interval for the true population proportion of people with kids. =![(0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)](/tpl/images/0391/1640/14892.png)
Ответ:
Shania is correct.
Step-by-step explanation:
For a right angled triangle, given two sides and an angle, we can use the sine rule or the various trigonometry identities to solve the triangle;
a/sin A = b/sinB
and
sin (angle) = opposite side/hypotenuse
cos (angle) = adjacent side/hypotenuse
tan (angle) = opposite/ adjacent
assuming a is the opposite side of angle A and b is also opposite of angle B
Otherwise, the only way to solve a triangle given two sides only is by the use of the Pythagoras theorem.
Hence, Shania is correct.