A random sample of married people was asked "Would you remarry your spouse if you were given the opportunity for a second time?" Of the 150 people surveyed, 127 of them said that they would do so. Find a 95% confidence interval for the proportion of married people who would remarry their spouse.a. 0.847 plusorminus 0.002b. 0.087 plusorminus 0.058c. 0.087 plusorminus 0.029d. 0.847 plusorminus 0.058

Answer:d. [tex]0.847 \pm 0.058[/tex]Step-by-step explanation:1) Notation and definitions[tex]X=127[/tex] number of people that would remarry.[tex]n=150[/tex] random sample taken[tex]\hat p=\frac{127}{150}=0.847[/tex] estimated proportion of people that would remarry.[tex]p[/tex] true population proportion of people that would remarry.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{\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 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]0.847 - 1.96\sqrt{\frac{0.847(1-0.847)}{150}}=0.789[/tex][tex]0.847 + 1.96\sqrt{\frac{0.847(1-0.847)}{150}}=0.905[/tex]The 95% confidence interval would be given by (0.789;0.905) and the margin of error is given by:[tex]Me=1.96\sqrt{\frac{0.847(1-0.847)}{150}}=0.058[/tex]So the best option for this case would be:d. [tex]0.847 \pm 0.058[/tex]