Respuesta :
Answer:
The 90% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.369, 0.431).
Step-by-step explanation:
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]1-\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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 675, \pi = \frac{270}{675} = 0.4[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.645\sqrt{\frac{0.4*0.6}{675}} = 0.369[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.645\sqrt{\frac{0.4*0.6}{675}} = 0.431[/tex]
The 90% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.369, 0.431).
Answer:
[tex]0.4 - 1.64\sqrt{\frac{0.4(1-0.4)}{675}}=0.369[/tex]
[tex]0.4 + 1.64\sqrt{\frac{0.4(1-0.4)}{675}}=0.431[/tex]
The 90% confidence interval would be given by (0.369;0.431)
Step-by-step explanation:
Previous concepts
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
Solution to the problem
The estimated proportion for this case is given by:
[tex] \hat p = \frac{x}{n}= \frac{270}{675}= 0.4[/tex]
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 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula: Â
[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]0.4 - 1.64\sqrt{\frac{0.4(1-0.4)}{675}}=0.369[/tex]
[tex]0.4 + 1.64\sqrt{\frac{0.4(1-0.4)}{675}}=0.431[/tex]
The 90% confidence interval would be given by (0.369;0.431)
