Chapter 7, Problem 55SE

a.

To determine

Find the standard error for each of the three firms given a sample of size 50 using the finite population correction factor.

Answer to Problem 55SE

The standard error for $N=2,000$ is 20.11.

The standard error for $N=5,000$ is 20.26.

The standard error for $N=10,000$ is 20.31.

Explanation of Solution

Calculation:

The given information is that sample of 50 items from its inventory is taken. Firm A’s inventory contains 2,000 items, firm B’s inventory contains 5,000 items and firm C’s inventory contains 10,000 items. The population standard deviation for the cost of item’s in each firm’s inventory is $\sigma =144$.

Sampling distribution of $\overline{x}$:

The probability distribution all possible values of the sample mean $\overline{x}$ is termed as the sampling distribution of $\overline{x}$.

• The expected value of $\overline{x}$ is, $E\left(\overline{x}\right)=\mu$.
• The standard deviation of $\overline{x}$ is

  For finite population, ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$

  For infinite population, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

• When the population size is infinite or finite and $\frac{n}{N}\le 0.05$ then the standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

For $N=2,000$:

The standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Substitute $\sigma$ as 144, N as 2,000 and n as 50 in the formula,

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)\\ =\sqrt{\frac{2,000-50}{2,000-1}}\left(\frac{144}{\sqrt{50}}\right)\\ =\sqrt{\frac{1,950}{1,999}}\left(\frac{144}{7.0711}\right)\\ =0.9877\left(20.3646\right)\end{array}$

     $=20.1141$

Thus, the standard deviation is 20.11.

For $N=5,000$:

The standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Substitute $\sigma$ as 144, N as 5,000 and n as 50 in the formula,

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)\\ =\sqrt{\frac{5,000-50}{5,000-1}}\left(\frac{144}{\sqrt{50}}\right)\\ =\sqrt{\frac{4,950}{4,999}}\left(\frac{144}{7.0711}\right)\\ =0.9951\left(20.3646\right)\end{array}$

     $=20.2648$

Thus, the standard deviation is 20.26.

For $N=10,000$:

The standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Substitute $\sigma$ as 144, N as 10,000 and n as 50 in the formula,

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)\\ =\sqrt{\frac{10,000-50}{10,000-1}}\left(\frac{144}{\sqrt{50}}\right)\\ =\sqrt{\frac{9,950}{9,999}}\left(\frac{144}{7.0711}\right)\\ =0.9975\left(20.3646\right)\end{array}$

     $=20.3137$

Thus, the standard deviation is 20.31.

b.

To determine

Find the probability that each firm the sample mean $\overline{x}$ will be within $\pm 25$ of the population mean $\mu$.

Answer to Problem 55SE

The probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ for $N=2,000$ is 0.785.

The probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ for $N=5,000$ is 0.7814.

The probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ for $N=10,000$ is 0.7814.

Explanation of Solution

Calculation:

For $N=2,000$:

The probability that the sample mean $\overline{x}$ will be within $\pm 25$ of the population mean $\mu$ is $P\left(\overline{x}\le \mu \pm 25\right)$

$\begin{array}{c}P\left(\overline{x}\ge \mu \pm 25\right)=P\left(\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right)\\ =P\left(\frac{-25}{20.11}\le z\le \frac{25}{20.11}\right)\\ =P\left(-1.24\le z\le 1.24\right)\\ =P\left(z\le 1.24\right)-P\left(z\le -1.24\right)\end{array}$

From Table 1: Cumulative probabilities for the standard normal distribution,

For $z=1.24$:

• Locate the value 1.2 in the first column.
• Locate the value 0.04 in the first row corresponding to the value 1.2 in the first column.
• The intersecting value of row and column is 0.8925.

For $z=-1.24$:

• Locate the value –1.2 in the first column.
• Locate the value 0.04 in the first row corresponding to the value –1.2 in the first column.
• The intersecting value of row and column is 0.1075.

$\begin{array}{c}P\left(\overline{x}\le \mu \pm 25\right)=P\left(z\le 1.24\right)-P\left(z\le -1.24\right)\\ =0.8925-0.1075\\ =0.785\end{array}$

Thus, the probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ is 0.785.

For $N=5,000$:

The probability that the sample mean $\overline{x}$ will be within $\pm 25$ of the population mean $\mu$ is $P\left(\overline{x}\le \mu \pm 25\right)$

$\begin{array}{c}P\left(\overline{x}\ge \mu \pm 25\right)=P\left(\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right)\\ =P\left(\frac{-25}{20.26}\le z\le \frac{25}{20.26}\right)\\ =P\left(-1.23\le z\le 1.23\right)\\ =P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\end{array}$

From Table 1: Cumulative probabilities for the standard normal distribution,

For $z=1.23$:

• Locate the value 1.2 in the first column.
• Locate the value 0.03 in the first row corresponding to the value 1.2 in the first column.
• The intersecting value of row and column is 0.8907.

For $z=-1.23$:

• Locate the value –1.2 in the first column.
• Locate the value 0.03 in the first row corresponding to the value –1.2 in the first column.
• The intersecting value of row and column is 0.1093.

$\begin{array}{c}P\left(\overline{x}\le \mu \pm 25\right)=P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\\ =0.8907-0.1093\\ =0.7814\end{array}$

Thus, the probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ is 0.7814.

For $N=10,000$:

The probability that the sample mean $\overline{x}$ will be within $\pm 25$ of the population mean $\mu$ is $P\left(\overline{x}\le \mu \pm 25\right)$.

$\begin{array}{c}P\left(\overline{x}\ge \mu \pm 25\right)=P\left(\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right)\\ =P\left(\frac{-25}{20.26}\le z\le \frac{25}{20.26}\right)\\ =P\left(-1.23\le z\le 1.23\right)\\ =P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\end{array}$

From Table 1: Cumulative probabilities for the standard normal distribution,

For $z=1.23$:

• Locate the value 1.2 in the first column.
• Locate the value 0.03 in the first row corresponding to the value 1.2 in the first column.
• The intersecting value of row and column is 0.8907.

For $z=-1.23$:

• Locate the value –1.2 in the first column.
• Locate the value 0.03 in the first row corresponding to the value –1.2 in the first column.
• The intersecting value of row and column is 0.1093.

$\begin{array}{c}P\left(\overline{x}\le \mu \pm 25\right)=P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\\ =0.8907-0.1093\\ =0.7814\end{array}$

Thus, the probability that the sample mean will be within $\pm 25$ of the population mean $\mu$ is 0.7814.

Here, all the probabilities are approximately same therefore for all three firms the sample of size 50 works.