In a recent poll, 370 people were asked if they liked dogs, and 7% said they did. Find the margin of error of this poll, at the 90% confidence level. Give your answer to three decimals.

Answer:[tex] ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218[/tex] Step-by-step explanation: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".  [tex]p[/tex] represent the real population proportion of interest [tex]\hat p=0.07[/tex] represent the estimated proportion for the sample n=370 is the sample size required (variable of interest) [tex]z[/tex] represent the critical value for the margin of error The population proportion have the following distribution  [tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/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=0.10[/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 margin of error for the proportion interval is given by this formula:  [tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)  And replacing into formula (a) the values provided we got:[tex] ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218[/tex] The margin of error on this case would be ME=0.0218