Respuesta :
Answer:
a) The 99% confidence interval would be given (0.601;0.679).
b) The 95% confidence interval would be given (0.611;0.669).
c) For part a:
[tex]Me= 2.58 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.0386[/tex]
For part b
[tex]Me= 1.96 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.0293[/tex]
d) Without computing the 90 % confidence interval we can conclude that the margin of error would be lower for the 90% confidence because the quantile [tex]z_{\alpha/2}[/tex] would be smaller compared to the quantiles for the 95% and 99% of confidence.
The 90% confidence interval would be given (0.615;0.665).
Step-by-step explanation:
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{\hat p(1-\hat p)}{n}})[/tex]
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]
[tex]\hat p =\frac{658}{1028}=0.64[/tex] estimated proportion of adults believe that marijuana should be legalized
n=1028 represent th sample size
Part a
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]
And replacing into the confidence interval formula we got:
[tex]0.64 - 2.58 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.601[/tex]
[tex]0.64 + 2.58 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.679[/tex]
And the 99% confidence interval would be given (0.601;0.679).
Part b
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.64 - 1.96 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.611[/tex]
[tex]0.64 + 1.96 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.669[/tex]
And the 95% confidence interval would be given (0.611;0.669).
Part c
The margin of error is given by:
[tex]Me= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For part a:
[tex]Me= 2.58 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.0386[/tex]
For part b
[tex]Me= 1.96 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.0293[/tex]
Part d
Without computing the 90 % confidence interval we can conclude that the margin of error would be lower for the 90% confidence because the quantile [tex]z_{\alpha/2}[/tex] would be smaller compared to the quantiles for the 95% and 99% of confidence.
For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex]0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.615[/tex]
[tex]0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{1028}}=0.665[/tex]
And the 90% confidence interval would be given (0.615;0.665).
