Question

The number of births per woman in China is $1.6$down from $5.91$in $1966$. This fertility rate has been attributed to the law passed in $1979$restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than $0.75$. They asked $50$women across China the number of births they had had. The results are shown in Table. Does the students’ survey indicate that the standard deviation is greater than $0.75$?

$\begin{array}{|ll|}\hline \text{\# of births}& \text{Frequency}\\ 0& 5\\ 1& 30\\ 2& 10\\ 3& 5\\ \hline\end{array}$

Short answer

We do not reject the null hypothesis because the $p$-value is greater than the level of significance.

Step-by-step solution

Given Information

It is given that the number of births per woman in China is $1.6$down from $5.91$in 1966 and the data for the number of births by $50$women is given.

Test whether the standard deviation of the number of births per woman in China is not greater than $0.75$or not.

The null and alternative hypotheses are:

$H_0:\sigma =0.75$

$H_1:\sigma >0.75$

Therefore, the expected number of students to attend their graduation is $19$.

Explanation

The calculation of mean and standard deviation is as shown below:

$\begin{array}{|cccc|}\hline \text{\# of births}\left(x\right)& \text{Frequency}\left(f\right)& x\times f& x^2\times f\\ 0& 5& 0& 0\\ 1& 30& 30& 30\\ 2& 10& 20& 40\\ 3& 5& 15& 45\\ \hline\end{array}$

Therefore, the mean and standard deviation will be calculated as:

$\overline{x}=\frac{\sum fx}{n}$

$=\frac{30+20+15}{5+30+10+5}$

$=1.3$

$s=\frac{\sum x^{2}f}{n}-{\overline{x}}^2$

$=\frac{30+40+45}{5+30+10+5}-1.3^2$

$=0.61$

The formula and calculation of the test statistic is:

${\chi }^2=\frac{(n-1)s^2}{{\sigma }^2}$

$=\frac{(50-1)0.{61}^2}{0.{75}^2}$

$=32.41$

The degrees of freedom are $50-1=49$. The formula and calculation of the $p$-value in Excel is:

Hence the$p$-value is$0.9675$