Question

7.59 Gestation Periods of Humans. For humans, gestation periods are normally distributed with a mean of 266 days and a standard deviation of 16 days. Suppose that you observe the gestation periods for a sample of nine humans.
a. Theoretically, what are the mean and standard deviation of all possible sample means?
b. Use the technology of your choice to simulate 2000 samples of nine human gestation periods each.
c. Determine the mean of each of the 2000 samples you obtained in part (b).
d. Roughly what would you expect the mean and standard deviation of the 2000 sample means you obtained in part (c) to be? Explain your answers.
e. Determine the mean and standard deviation of the 2000 sample means you obtained in part (c).
f. Compare your answers in parts (d) and (e). Why are they different?

Short answer

(a) The mean is $266$, and the standard Deviation is $4.3333$.

(b) Depending on the initial number chosen, the solution may differ.
(c) Depending on the initial number chosen, the solution may differ.
(d) The mean $=266$, and the standard Deviation$=4.3333$.
(e) Depending on the initial number chosen, the solution may differ.

(f) The main variation between the results is due to sampling error in parts (d) and (e).

Step-by-step solution

Part (a) Step 1: Given information

To determine the mean and standard deviation of all possible sample means.

Part (a) Step 2: Explanation

Let the given values are:

Mean is $\mu =6$

Standard deviation is $\sigma =16$

And the sample size is $\mathrm{n}=9$

The mean of a sample's sample means of distribution is known to be the same as the population mean.

${\mu }_{\overline{x}}=\mu$

$=266$

The standard deviation of the sample means is equal to the population standard deviation divided by the square root of the sample size.

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

$=\frac{16}{\sqrt{9}}$$=\frac{16}{\sqrt{9}}$

$=\frac{16}{3}$

$\approx 4.3333$

As a result, the mean is $266$and the standard Deviation is $4.3333$.

Part (b) Step 1: Given information

To use the technology of own choice to simulate $2000$ samples of nine human gestation periods each.

Part (b) Step 2: Explanation

A normal distribution of data values can be used to imitate the data values.
Let, the mean $=266$
And the standard Deviation $=4.3333$
Using the following formula in Excel:
$=NORMINV(RAND(;266;16)$
The above command can be used to replicate each of the $9$sample values, and then the process can be continued until $2000$samples with $9$ values have been calculated.
The following is a snapshot of the final result:

273.617729728	266.477776441	289.722462117	269.95608791	238.104384977
257.737065784	260.19981211	254.402305358	269.092995231	256.953301088
300.268577649	287.399240314	290.450140962	242.775036729	253.642757669
250.831786839	264.938704095	257.631014766	271.626418278	275.155829109
252.164456852	248.644850741	261.586877048	273.78111226	262.481219206
234.185128179	252.588501472	255.463755401	274.363659553	296.691440508
252.63756912	227.64873443	266.11317732	277.417687018	255.602895117
239.224834824	261.668050501	288.487020264	305.193006855	259.987114748
263.749098399	259.162529914	268.785944817	255.48561214	284.430682465

As a result, depending on the initial number chosen, the solution may differ.

Part (c) Step 1: Given information

To determine the mean of each of the $2000$ samples that obtained in part (b).

Part (c) Step 2: Explanation

The mean can be computed by multiplying the total of all values by the number of values.
For the third column, for example, the following calculation is performed:
$\overline{x}=\frac{289.72+\dots +268.79}{9}$

$=\frac{2432.64}{9}$

$=270.29$

This can be done for each of the columns.

As a result, depending on the initial number chosen, the solution may differ.

Part (d) Step 1: Given information

To determine the mean and standard deviation of the $2000$ sample means that obtained in part (c).

Part (d) Step 1: Explanation

The expected result as follows:
The mean$=266$
And the standard Deviation $=4.3333$
The method of reasoning for this is that the distribution of the $2000$sample means will be comparable to the sampling distribution of the sample mean.

Part (e) Step 1: Given information

To determine the mean and standard deviation of the $2000$ sample means that obtained in part (c).

Part (e) Step 2: Explanation

The mean can be computed by multiplying the total of all values by the number of values.
$\overline{x}=\frac{289.72+\dots +268.79}{2000}$

$=266$

$s=\sqrt{\frac{(289.72-266{)}^2+\dots +(268.79-266{)}^2}{2000-1}}$

$\approx 4.3052$

As a result, depending on the initial number chosen, the solution may differ.

Part (f) Step 1: Given information

To compare the answers in parts (d) and (e). And find the difference between parts (d) and (e).

Part (e) Step 2: Explanation

The main variation between the results is due to sampling error in parts (d) and (e).
To be more specific, even though the numbers are fairly near to the expected values, various samples will have different means and standard deviations.