91影视

The number of birth in New York State is recorded.

The number of births is $n=860$with 426 boys.

The confidence interval is$95\%$

The requirements are verified as follows,

1. The samples are selected randomly and normally distributed.
2. There are two categories of outcomes, either boy or girl.
3. The counts are computed as follows,

$$\begin{gathered} np=860脳0.512 \\ =440.32 \\ >0.05 \end{gathered}$$

And

$$\begin{gathered} nq=860脳0.488 \\ =419.68 \\ >0.05 \end{gathered}$$

All the conditions are satisfied. Hence the 95% confidence interval for population proportion can be estimated using the z-test.

Thesample proportion of the boys is:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{426}{860} \\ =0.495 \end{gathered}$$

Therefore, the sample proportion is 0.495.

Then,

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.495 \\ =0.505 \end{gathered}$$

At confidence interval, $伪=0.05$.

Thus, using the standard normal table,

$$\begin{gathered} z_{crit}=z_{\frac{伪}{2}} \\ =1.96 \end{gathered}$$

The margin of error is given by,

$$\begin{gathered} E=z_{crit}脳\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =1.96脳\sqrt{\frac{0.495脳0.505}{860}} \\ =0.0334 \end{gathered}$$

The margin of error is 0.0334.

The formula for the confidence interval is given by,

$$\begin{gathered} CI=\hat{p}-E<p<\hat{p}+E \\ =\left(0.495-0.0334<p<0.495+0.0334\right) \\ =(0.4616<p<0.5284) \end{gathered}$$

Thus, 95% confidence interval is between 0.462 to 0.528.

The confidence interval is from 0.462 to 0.528. The population proportion 0.512 is included in the interval. Hence there is astrong evidence in support of the claim.