Chapter 7: Q. 7.48 (page 301)

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 $\bar{x}$ . Interpret your results in words.
b. Repeat part (a) with $n = 120$ .

Short Answer

Expert verified

(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 $x$ , for the sample size of $40$ .

Part (a) Step 2: Explanation

Determine the standard deviation as:
$σ_{\bar{x}} = \frac{σ}{\sqrt{n}}$
The mean $(μ)$ for the age of surgical or natural, is $44.8$ years.
$μ_{x} = μ$

$μ_{x} = 44.8$

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

σ_{\bar{x}} = \frac{σ}{\sqrt{n}}

σ_{\bar{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:
$σ_{\bar{x}} = \frac{σ}{\sqrt{n}}$
The mean $(μ)$ for the age of surgical or natural, is $44.8$ years.
$μ_{x} = μ$
$μ_{x} = 44.8$
Determine the standard deviation by substitute $5.87$ for $σ$ and $120$ for $n$ in the formula of the standard deviation.
$σ_{\bar{x}} = \frac{σ}{\sqrt{n}}$

$σ_{\bar{x}} = \frac{5.87}{\sqrt{120}}$

= 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.

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Short Answer

Step by step solution

Part (a) Step 1: Given information

Part (a) Step 2: Explanation

Part (b) Step 1: Given information

Part (b) Step 2: Explanation

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