Respuesta :
Answer:
a) [tex]\hat p = \frac{X}{n}=\frac{564}{916}= 0.616[/tex]
b) For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
The margin of error is given by:
[tex] ME= 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.0415[/tex]
c) [tex]0.616 - 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.575[/tex]
[tex]0.616 + 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.657[/tex]
And the 90% confidence interval would be given (0.575;0.657).
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
Part a
For this case the best point of estimate of the population proportion is:
[tex]\hat p = \frac{X}{n}=\frac{564}{916}= 0.616[/tex]
Part b
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
The margin of error is given by:
[tex] ME= 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.0415[/tex]
Part c
And replacing into the confidence interval formula we got:
[tex]0.616 - 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.575[/tex]
[tex]0.616 + 2.58 \sqrt{\frac{0.616(1-0.616)}{916}}=0.657[/tex]
And the 90% confidence interval would be given (0.575;0.657).
A) The best point estimate of the population proportion is; p^ = 0.6157
B) The value of the margin of error for the poll is;
E = 0.0414
C) The confidence interval for the given poll is;
CI = (0.5743, 0.6571)
We are given;
Number of people in the poll; n = 916
Number of people who said yes; x = 564
A) Population proportion would be gotten from the formula;
p^ = x/n
p^ = 564/916
p^ = 0.6157
B) The formula for the margin of error is;
E = (z_α/2)√(p^(1 - p^)/n)
Where;
(z_α/2) is the critical value at the given confidence level.
Now,the critical value z_α/2 for confidence level of 99% is gotten from tables to be 2.576.
Thus;
E = 2.576√(0.6157(1 - 0.6157)/916)
E = 0.0414
C) Formula for the confidence interval is;
CI = p^ ± (z_α/2)√(p^(1 - p^)/n)
From B we saw that (z_α/2)√(p^(1 - p^)/n) = 0.0414
Thus;
CI = 0.6157 ± 0.0414
CI = (0.6157 - 0.0414), (0.6157 + 0.0414)
CI = (0.5743, 0.6571)
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