Nominal Data. In Exercises 9–12, use the sign test for the claim involving nominal data.

Births A random sample of 860 births in New York State included 426 boys and 434 girls. Use a 0.05 significance level to test the claim that when babies are born, boys and girls are equally likely.

The number of male births is equal to 426, and the number of female births is equal to 434.

The researcher wants to test that the boys and girls are equally likely to be born at the significance level of 0.05.

The sign test is the non-parametric test used to test the claim of difference between the proportions of male and female births.

Let p be the proportion of male births.

Considering that the proportions of male and female births should be the same, the null hypothesis is as follows:

\({H_0}:p = 0.5\)

Boys and girls are equally likely to be born.

The alternative hypothesis is as follows:

\({H_0}:p \ne 0.5\)

Boys and girls are not equally likely to be born.

The test is two-tailed.

A negative sign denotes the male births.

A positive sign denotes the female births.

The number of negative signs =426.

The number of positive signs =434.

The sample size (n) is equal to 860.

Let x be the number of times the less frequent sign occurs.

The less frequent sign is the negative sign corresponding to the number of male births.

The value of xis equal to 426.

As the sample size n is greater than 25, the value of z is calculated.

The test statistic z is calculated as shown:

\(\begin{array}{c}z = \frac{{\left( {x + 0.5} \right) - \frac{n}{2}}}{{\frac{{\sqrt n }}{2}}}\\ = \frac{{\left( {426 + 0.5} \right) - \frac{{860}}{2}}}{{\frac{{\sqrt {860} }}{2}}}\\ = - 0.24\end{array}\)

Critical value:

The critical value of z from the standard normal table for a two-tailed test with a value of\(\alpha \)= 0.05 is equal to\( \pm 1.96\).

Moreover, the absolute value of z equal to 0.24 is less than the critical value; the null hypothesis is failed to reject.

P-value:

The corresponding p-value for z-score equal to -0.24 and\(\alpha \)equal to 0.05 from the table is equal to 0.4052.

Since it is a two-tailed test, the p-value becomes as follows:

\(\begin{array}{c}p{\rm{ - value}} = 2 \times 0.4052\\ = 0.8104\end{array}\)

As the p-value equal to 0.8104 is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that boys and girls are equally likely to be born.