Mathematics, 21.12.2019 03:31 iixyloa

Is it plausible that more than 10% of americans believe in aliens? a random sample of 2000 adult americans were surveyed and 15% of them said that they believed in aliens. find the 95% confidence interval for the proportion of americans who believe in aliens then choose the correct interpretation. (round to the nearest tenth of a percent) a. the population proportion of americans who believe in aliens is between 10% +/- 1.6% with a confidence level of 95%. the interval includes 10% and therefore, it is plausible that at least 10% of americans believe in alien fin b. the population proportion of americans who believe in aliens is between 15% +/- 0.8% with a confidence level of 95%. the interval does not include 10% and therefore, it is not plausible that at least 10% of americans believe in aliens. 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.

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Answer from: maddy121209

The 95% confidence interval would be given by (0.134;0.166)

$ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156$ 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

number of Americans that said that they believed in aliens

n=2000 random sample taken

$\hat p=0.15$ estimated proportion of Americans that said that they believed in aliens

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

$p \sim N(p,\sqrt{\frac{p(1- p)}{n}})$

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 $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.15 - 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.134$

$0.15 + 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.166$

The 95% confidence interval would be given by (0.134;0.166)

The margin of error is given by:

$ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156$ and that correspond with approximately 1.6% of margin of error

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Answer from: Quest

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