Respuesta :
Answer:
a) [tex]0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487[/tex]
[tex]0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553[/tex]
The 95% confidence interval would be given by (0.487;0.553)
b) After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes.
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".
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".
Solution to the problem
Part a
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]
If we replace the values obtained we got:
[tex]0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487[/tex]
[tex]0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553[/tex]
The 95% confidence interval would be given by (0.487;0.553)
Part b
After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes at 5% of significance.
