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A sample of 860 births is selected, out of which 426 areboys. It is claimed that 51.2% of newborn babies are boys.

The null hypothesis is written as follows.

The proportion of newborn babies that are boys is equal to 51.2%.

\({H_0}:p = 0.512\)

The alternative hypothesis is written as follows.

The proportion of newborn babies that are boys is not equal to 51.2%.

\({H_1}:p \ne 0.512\)

The test is two-tailed.

The sample size is n=860.

The sample proportion of babies that are boys is computed below.

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{boys}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{babies}}\;}}\\ = \frac{{426}}{{860}}\\ = 0.495\end{array}\].

The population proportion of babies that are boys is equal to 0.512.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.495 - 0.512}}{{\sqrt {\frac{{0.512\left( {1 - 0.512} \right)}}{{860}}} }}\\ = - 0.977\end{array}\).

Thus, z=-0.977.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of -0.977 is equal to 0.3286.

Asthe p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of babies that are boys is equal to 51.2%.

The results support the belief that 51.2% of the babies born are boys.