Respuesta :
Answer:
i) [tex]ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And replacing we got:
[tex] ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392[/tex]
ii) [tex] Lower = 0.8-0.0392= 0.7608[/tex]
[tex] Upper = 0.8 +0.0392= 0.8392[/tex]
Step-by-step explanation:
Notation and definitions
[tex]X=320[/tex] number of people that claimed always buckle up.
[tex]n=400[/tex] random sample taken
[tex]\hat p=\frac{320}{400}=0.8[/tex] estimated proportion of people that claimed always buckle up
[tex]p[/tex] true population proportion of people that claimed 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".
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]
Solution to the problem
Part i
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]
The margin of error is given by:
[tex]ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And replacing we got:
[tex] ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392[/tex]
Part ii
And the confidence interval would be given by:
[tex] Lower = 0.8-0.0392= 0.7608[/tex]
[tex] Upper = 0.8 +0.0392= 0.8392[/tex]
From the information given, it is found that:
i) The margin of error is of 0.0392.
ii) The confidence interval is (0.7608, 0.8392).
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have that [tex]n = 400, \pi = \frac{320}{400} = 0.8[/tex].
95% confidence level
So [tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
Hence, the margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.8(0.2)}{400}}[/tex]
[tex]M = 0.0392[/tex]
The margin of error is of 0.0392.
Then, the interval is:
[tex]\pi - M = 0.8 - 0.0392 = 0.7608[/tex]
[tex]\pi + M = 0.8 + 0.0392 = 0.8392[/tex]
The confidence interval is (0.7608, 0.8392).
A similar problem is given at https://brainly.com/question/16807970
