Suppose a random sample of size 50 is selected from a population with . Find the value of the standard error of the
mean in each of the following cases. (Use the finite population correction factor if appropriate. Round your answers to two
decimal places.)
(a) The population size is infinite.
(b) The population size is .
© The population size is .
(d) The population size is .

Ask by Hanson Little. in the United States

Mar 16,2025

Question

Suppose a random sample of size 50 is selected from a population with . Find the value of the standard error of the
mean in each of the following cases. (Use the finite population correction factor if appropriate. Round your answers to two
decimal places.)
(a) The population size is infinite.
(b) The population size is .
© The population size is .
(d) The population size is .

Ask by Hanson Little. in the United States

Mar 16,2025

UpStudy AI · Accepted Answer

We are given a sample size $$ n = 50 $$ and a population standard deviation $$ \sigma = 8 $$. The standard error (SE) of the mean for a finite population is given by

$$
SE = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N - n}{N - 1}},
$$

where $$ N $$ is the population size. When the population is infinite, the finite population correction factor is not used and the SE is simply

$$
SE = \frac{\sigma}{\sqrt{n}}.
$$

We now work through each part.

---

**(a) Population size is infinite**

When $$ N $$ is infinite, no correction factor is needed. Thus

$$
SE = \frac{8}{\sqrt{50}}.
$$

Calculate

$$
\sqrt{50} \approx 7.07,
$$

so

$$
SE \approx \frac{8}{7.07} \approx 1.13.
$$

---

**(b) Population size is $$ N = 50,\!000 $$**

Here the finite population correction factor is

$$
\sqrt{\frac{N - n}{N - 1}} = \sqrt{\frac{50,\!000 - 50}{50,\!000 - 1}} = \sqrt{\frac{49,\!950}{49,\!999}}.
$$

Since $$ \frac{49,\!950}{49,\!999} $$ is very close to $$ 1 $$, the factor is approximately $$ 1 $$:

$$
\sqrt{\frac{49,\!950}{49,\!999}} \approx 0.999.
$$

Thus

$$
SE \approx \frac{8}{\sqrt{50}} \times 0.999 \approx 1.13 \times 0.999 \approx 1.13.
$$

---

**(c) Population size is $$ N = 5,\!000 $$**

Now the correction factor is

$$
\sqrt{\frac{5,\!000 - 50}{5,\!000 - 1}} = \sqrt{\frac{4,\!950}{4,\!999}}.
$$

We compute

$$
\frac{4,\!950}{4,\!999} \approx 0.9900,
$$

and therefore

$$
\sqrt{0.9900} \approx 0.995.
$$

Thus

$$
SE \approx \frac{8}{\sqrt{50}} \times 0.995 \approx 1.13 \times 0.995 \approx 1.13.
$$

---

**(d) Population size is $$ N = 500 $$**

Here the correction factor is

$$
\sqrt{\frac{500 - 50}{500 - 1}} = \sqrt{\frac{450}{499}}.
$$

Calculate

$$
\frac{450}{499} \approx 0.902,
$$

and so

$$
\sqrt{0.902} \approx 0.95.
$$

Then

$$
SE \approx \frac{8}{\sqrt{50}} \times 0.95 \approx 1.13 \times 0.95 \approx 1.07.
$$

---

**Final Answers:**

- (a) $$ SE \approx 1.13 $$
- (b) $$ SE \approx 1.13 $$
- (c) $$ SE \approx 1.13 $$
- (d) $$ SE \approx 1.07 $$
The standard error of the mean is approximately 1.13 for parts (a), (b), and (c), and 1.07 for part (d).

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