Question

7.48 Menopause in Mexico. In the article "Age at Menopause in Puebla. Mexico" (Human Biology, Vol. 75, No, 2, Pp. 205-206), authors L. Sievert and S. Hautaniemi compared the age of menopause for different populations. Menopause, the last menstrual period, is a universal phenomenon among females. According to the article, the mean age of menopause, surgical or natural, in Puebla, Mexico is $44.8$years with a standard deviation of $5.87$years. Let $\tilde{x}$denote the mean age of menopause for a sample of females in Puebla, Mexico.
a. For samples of size $40$, find the mean and standard deviation of $\overline{x}$. Interpret your results in words.
b. Repeat part (a) with $n=120$.

Short answer

(a) The mean and standard deviation for sample size $40$is $44.8$years and $0.928$years respectively.

(b) The mean and standard deviation for sample size $120$is $44.8$years and $0.536$ years respectively.

Step-by-step solution

Part (a) Step 1: Given information

To find the standard deviation of $\overline{x}$, for the sample size of $40$.

Part (a) Step 2: Explanation

Determine the standard deviation as:
${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
The mean $\left(\mu \right)$for the age of surgical or natural, is $44.8$ years.
${\mu }_x=\mu$

${\mu }_x=44.8$

Determine the standard deviation by substitute $5.87$for $\sigma$and $40$for $n$in the formula of standard deviation.

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
${\sigma }_{\overline{x}}=\frac{5.87}{\sqrt{40}}$
$=\frac{5.87}{6.3246}$
$=0.928$
As a result, the mean and standard deviation of all possible sample mean age of menopause, surgical or natural, are $44.8$years and $0.928$years, respectively, for samples of size $40$ women.

Part (b) Step 1: Given information

To find the mean and standard deviation for sample size $120$.

Part (b) Step 2: Explanation

Determine the standard deviation as:
${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
The mean $\left(\mu \right)$for the age of surgical or natural, is $44.8$ years.
${\mu }_x=\mu$
${\mu }_x=44.8$
Determine the standard deviation by substitute $5.87$for $\sigma$and $120$for $n$in the formula of the standard deviation.
${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

${\sigma }_{\overline{x}}=\frac{5.87}{\sqrt{120}}$

$=0.536$
As a result, the mean and standard deviation of all potential sample mean age of menopause, surgical or natural, are $44.8$years and $0.536$ years, respectively, for samples of size $40$ women.