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Answer:
The 95% confidence interval would be given by (0.134;0.166)
[tex]ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156[/tex] and that correspond with approximately 1.6% of margin of error
C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.
Step-by-step explanation:
Notation and definitions
[tex]X[/tex] number of Americans that said that they believed in aliens
[tex]n=2000[/tex] random sample taken
[tex]\hat p=0.15[/tex] estimated proportion of Americans that said that they believed in aliens
[tex]p[/tex] true population proportion of Americans that said that they believed in aliens
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1- p)}{n}})[/tex]
Confidence interval
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.15 - 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.134[/tex]
[tex]0.15 + 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.166[/tex]
The 95% confidence interval would be given by (0.134;0.166)
The margin of error is given by:
[tex]ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156[/tex] and that correspond with approximately 1.6% of margin of error
C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.
The random sample of the Americans show that C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%.
How to illustrate the confidence interval?
From the information given, the confidence interval will be computed as having 0.134 and 0.1657 respectively.
Also, the margin of error will be:
= 1.96 × [✓0.15(1 - 0.15)/2000
= 0.0156
Therefore, the population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.
Learn more about confidence interval on:
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