In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with 41% of the participants saying they start in this position. Use a normal distribution to find a 99% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016. Round your answers to three decimal places. The 99% confidence interval is Enter your answer; The 95% confidence interval, value 1 to Enter your answer; The 95% confidence interval, value 2 .

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]0.41 - 1.96\sqrt{\frac{0.41(1-0.41)}{1000}}=0.380[/tex]

[tex]0.41 + 1.96\sqrt{\frac{0.41(1-0.41)}{1000}}=0.440[/tex]

The 95% confidence interval would be given by (0.380;0.440)

And for the 99% confident interval the critical value would be 2.58 and if we replace we got:

[tex]0.41 - 2.58\sqrt{\frac{0.41(1-0.41)}{1000}}=0.370[/tex]

[tex]0.41 + 2.58\sqrt{\frac{0.41(1-0.41)}{1000}}=0.450[/tex]

The 99% confidence interval would be given by (0.370;0.450)