Question

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 1

a. 7700, (.7321, .8079)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

Z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

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
`n = 474, \pi = (365)/(474) = 0.77`. The point estimate is 0.77.

95% confidence interval

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.77 - 1.96\sqrt{(0.77*0.23)/(474)} = 0.7321`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.77 + 1.96\sqrt{(0.77*0.23)/(474)} = 0.8079`

The correct answer is:

a. 7700, (.7321, .8079)