Question

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a confidence interval estimate of p, then address the given question.

BirthsA random sample of 860 births in New York State included 426 boys. Construct a 95% confidence interval estimate of the proportion of boys in all births. It is believed that among all births, the proportion of boys is 0.512. Do these sample results provide strong evidence against that belief?

Short answer

a. The 95% confidence interval is between0.462 and 0.528

b. There is strong evidence for the belief.

Step-by-step solution

Given information

The number of birth in New York State is recorded.

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

The confidence interval is$95\%$

Check the requirement

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,

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

And

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

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

Calculate the sample proportion

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,

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

Compute the critical value

At confidence interval, $\alpha =0.05$.

Thus, using the standard normal table,

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

Compute margin of error

The margin of error is given by,

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

The margin of error is 0.0334.

Compute the confidence interval

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.

Conclusion

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.