La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of$21$ U.S. mothers who recently weaned their children. The mean weaning age was nine monthsrole="math" localid="1649394170256" $(3/4\text{year})$ with a standard deviation of $4$ months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

$M=9\text{months},S=4\text{months},n=21,a=0.05$

Hypothesis: The null hypothesis states that in the United States, the average weaning age is greater than or equal to four years, or $48$months. According to the alternative hypothesis, the average age of weaning in the United States is less than four years or $48$months. The hypothesis can be represented as:

$H_0:\mu \ge 48$

$H_a:\mu <4$

The degree of freedom is calculated by:

$$\begin{gathered} df=n-1 \\ =21-1 \\ =20 \end{gathered}$$

From the t distribution table, for a left tailed with $a=0.05$for $df=20$, the critical value of t values for the critical region are $t=-1.725$.

The test statistic is:

localid="1649831515650" $$\begin{gathered} t=\frac{M-\mu }{{\displaystyle \frac{S}{\sqrt{n}}}} \\ \text{whereSisthestandarddeviation,nisthesamplesize} \\ S=4,n=21 \\ =\frac{9-48}{{\displaystyle \frac{4}{\sqrt{21}}}} \\ =\frac{-39}{0.8729} \\ =-44.68 \end{gathered}$$

Since the null hypothesis is rejected, the average age of weaning in the United States is less than $4$years or$48$ months.