Dr. Cyril conducts a simple random sample of 500 men who became fathers for the first time in the past year. He finds that 23% of them report being unsure of their ability to be good fathers, plus or minus 4%. What does this mean?

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{p(1-p)}{n}})[/tex]

Solution to the problem

For this case we have the following datast given:

[tex]X=500*0.23=115[/tex] number of men who became fathers for the first time in the past year

[tex]n=500[/tex] random sample taken

[tex]\hat p=\frac{115}{500}=0.23[/tex] estimated proportion of men who became fathers for the first time in the past year.

[tex]p [/tex] true population proportion of men who became fathers for the first time in the past year

The confidence interval for the proportion is given by the following formula:

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

And for this case we knwow that the margin of error is given by this formula:

[tex] ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

And based on the info given we have that:

[tex] ME = 4\% = 0.04 = z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

So then we can find the limits for the interval like this:

[tex] Lower = 0.23-0.04 = 0.19[/tex]

[tex] Upper = 0.23+0.04 = 0.27[/tex]

We can conclude that the true percentage of men who became fathers for the first time in the past year is between 19% and 27%