In a survey of 474 U.S. women, 365 said that the media has a negative effect on women's health because they set unattainable standards for appearance. Let p denote the proportion of all women who think media has a negative impact on women's health because they set unattainable standards for appearance. Find a point estimate for p and also construct a 95% confidence interval for p. .a. 7700, (.7321, .8079)b .2300, (.1921, -2679)c. 365, (.7683, .7717)d. 365, (.7507, .7893)e .7700, (.7507, .7893).

Answer:a. 7700, (.7321, .8079)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 interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]In whichZ is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].For this problem, we have that:In a survey of 474 U.S. women, 365 said that the media has a negative effect on women's health because they set unattainable standards for appearance. This means that [tex]n = 474, \pi = \frac{365}{474} = 0.77[/tex]. The point estimate is 0.77.95% confidence intervalSo [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].The lower limit of this interval is:[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.96\sqrt{\frac{0.77*0.23}{474}} = 0.7321[/tex]The upper limit of this interval is:[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.96\sqrt{\frac{0.77*0.23}{474}} = 0.8079[/tex]The correct answer is:a. 7700, (.7321, .8079)