Respuesta :
Answer:
The 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex] Â
Step-by-step explanation:
1) Data given and notation Â
n=400 represent the random sample taken  Â
X=320 represent the people drivers claimed they always buckle up
[tex]\hat p=\frac{320}{400}=0.8[/tex] estimated proportion of people drivers claimed they always buckle up
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed) Â Â
Confidence =95% or 0.95
p= population proportion of people drivers claimed they always buckle up
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
2) Calculating the interval for the proportion
The confidence interval would be given by this formula Â
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex] Â
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. Â
[tex]z_{\alpha/2}=1.96[/tex] Â
And replacing into the confidence interval formula we got: Â
[tex]0.8 - 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.761[/tex] Â
[tex]0.8 + 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.839[/tex] Â
And the 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex] Â
We are confident that about 76.1% to 83.9% of people drivers that they always buckle up  at 95% of confidence
