Chapter 7.6, Problem 30E

The population proportion is .30. What is the probability that a sample proportion will be within ±.04 of the population proportion for each of the following sample sizes?

1. a. n = 100
2. b. n = 200
3. c. n = 500
4. d. n = 1000
5. e. What is the advantage of a larger sample size?

To determine

Find the probability that the a sample proportion will be within $\pm 0.04$ of the population proportion for $n=100$.

Answer to Problem 30E

The probability that a sample proportion will be within $\pm 0.04$ of the population proportion for $n=100$ is 0.6156.

Explanation of Solution

Calculation:

The given information is that the population proportion is 0.30.

Sampling distribution of $\overline{p}$:

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

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

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

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

• When $np\ge 5$ and $n\left(1-p\right)\ge 5$ then the sampling distribution of $\overline{p}$ is approximated by a normal distribution.

The expected value of $\overline{p}$ is,

$\begin{array}{c}E\left(\overline{p}\right)=p\\ =0.30\end{array}$

Thus, the expected value of $\overline{p}$ is 0.30.

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

Substitute p as 0.30 and n as 100 in the formula,

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.30\left(1-0.30\right)}{100}}\\ =\sqrt{\frac{0.21}{100}}\\ =\sqrt{0.0021}\end{array}$

      $=0.0458$

Thus, the standard deviation of $\overline{p}$ is 0.0458.

Here,

$\begin{array}{c}np=100\left(0.30\right)\\ =30\\ \ge 5\end{array}$

$\begin{array}{c}n\left(1-p\right)=100\left(1-0.30\right)\\ =100\left(0.70\right)\\ =70\\ \ge 5\end{array}$

Therefore, the sampling distribution of $\overline{p}$ is approximately normal with $E\left(\overline{p}\right)=0.30$ and ${\sigma }_{\overline{p}}=0.0458$.

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=100$ is, $P\left(\overline{p}\le p\pm 0.04\right)=P\left(0.26\le \overline{p}\le 0.34\right)$

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left[\frac{0.26-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{\overline{p}-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{0.34-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\right]\\ =P\left(\frac{0.26-0.30}{0.0458}\le z\le \frac{0.34-0.30}{0.0458}\right)\\ =P\left(-0.87\le z\le 0.87\right)\\ =P\left(z\le 0.87\right)-P\left(z\le -0.87\right)\end{array}$

For $z=0.87$:

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

• Locate the value 0.8 in the first column.
• Locate the value 0.07 in the first row corresponding to the value 0.8 in the first column.
• The intersecting value of row and column is 0.8078.

For $z=-0.87$:

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

• Locate the value –0.8 in the first column.
• Locate the value 0.07 in the first row corresponding to the value –0.8 in the first column.
• The intersecting value of row and column is 0.1922.

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left(z\le 0.87\right)-P\left(z\le -0.87\right)\\ =0.8078-0.1922\\ =0.6156\end{array}$

Thus, the probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=100$ is 0.6156.

To determine

Find the probability that the a sample proportion will be within $\pm 0.04$ of the population proportion for $n=200$.

Answer to Problem 30E

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=200$ is 0.7814.

Explanation of Solution

Calculation:

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

Substitute p as 0.30 and n as 200 in the formula,

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.30\left(1-0.30\right)}{200}}\\ =\sqrt{\frac{0.21}{200}}\\ =\sqrt{0.00105}\end{array}$

      $=0.0324$

Thus, the standard deviation of $\overline{p}$ is 0.0324.

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=200$ is, $P\left(\overline{p}\le p\pm 0.04\right)=P\left(0.26\le \overline{p}\le 0.34\right)$

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left[\frac{0.26-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{\overline{p}-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{0.34-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\right]\\ =P\left(\frac{0.26-0.30}{0.0324}\le z\le \frac{0.34-0.30}{0.0324}\right)\\ =P\left(-1.23\le z\le 1.23\right)\\ =P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\end{array}$

For $z=1.23$:

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

• 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 $z=-1.23$:

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

• 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

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left(z\le 1.23\right)-P\left(z\le -1.23\right)\\ =0.8907-0.1093\\ =0.7814\end{array}$

Thus, the probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=200$ is 0.7814.

To determine

Find the probability that the a sample proportion will be within $\pm 0.04$ of the population proportion for $n=500$.

Answer to Problem 30E

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=500$ is 0.9488.

Explanation of Solution

Calculation:

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

Substitute p as 0.30 and n as 500 in the formula,

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.30\left(1-0.30\right)}{500}}\\ =\sqrt{\frac{0.21}{500}}\\ =\sqrt{0.00042}\end{array}$

      $=0.0205$

Thus, the standard deviation of $\overline{p}$ is 0.0205.

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=500$ is, $P\left(\overline{p}\le p\pm 0.04\right)=P\left(0.26\le \overline{p}\le 0.34\right)$

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left[\frac{0.26-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{\overline{p}-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{0.34-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\right]\\ =P\left(\frac{0.26-0.30}{0.0205}\le z\le \frac{0.34-0.30}{0.0205}\right)\\ =P\left(-1.95\le z\le 1.95\right)\\ =P\left(z\le 1.95\right)-P\left(z\le -1.95\right)\end{array}$

For $z=1.95$:

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

• Locate the value 1.9 in the first column.
• Locate the value 0.05 in the first row corresponding to the value 1.9 in the first column.
• The intersecting value of row and column is 0.9744.

For $z=-1.95$:

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

• Locate the value –1.9 in the first column.
• Locate the value 0.05 in the first row corresponding to the value –1.9 in the first column.
• The intersecting value of row and column is 0.0256.

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left(z\le 1.95\right)-P\left(z\le -1.95\right)\\ =0.9744-0.0256\\ =0.9488\end{array}$

Thus, the probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=500$ is 0.9488.

To determine

Find the probability that the a sample proportion will be within $\pm 0.04$ of the population proportion for $n=1,000$.

Answer to Problem 30E

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=1,000$ is 0.9942.

Explanation of Solution

Calculation:

The sample size is 1,000.

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

Substitute p as 0.30 and n as 1,000 in the formula,

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}.\\ =\sqrt{\frac{0.30\left(1-0.30\right)}{1,000}}\\ =\sqrt{\frac{0.21}{1,000}}\\ =\sqrt{0.00021}\end{array}$

      $=0.0145$

Thus, the standard deviation of $\overline{p}$ is 0.0145.

The probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=1,000$ is, $P\left(\overline{p}\le p\pm 0.04\right)=P\left(0.26\le \overline{p}\le 0.34\right)$

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left[\frac{0.26-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{\overline{p}-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\le \frac{0.34-E\left(\overline{p}\right)}{{\sigma }_{\overline{p}}}\right]\\ =P\left(\frac{0.26-0.30}{0.0145}\le z\le \frac{0.34-0.30}{0.0145}\right)\\ =P\left(-2.76\le z\le 2.76\right)\\ =P\left(z\le 2.76\right)-P\left(z\le -2.76\right)\end{array}$

For $z=2.76$:

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

• Locate the value 2.7 in the first column.
• Locate the value 0.06 in the first row corresponding to the value 2.7 in the first column.
• The intersecting value of row and column is 0.9971.

For $z=-2.76$:

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

• Locate the value –2.7 in the first column.
• Locate the value 0.06 in the first row corresponding to the value –2.7 in the first column.
• The intersecting value of row and column is 0.0029.

$\begin{array}{c}P\left(0.26\le \overline{p}\le 0.34\right)=P\left(z\le 2.76\right)-P\left(z\le -2.76\right)\\ =0.9971-0.0029\\ =0.9942\end{array}$

Thus, the probability that the sample proportion will be within $\pm 0.04$ of the population proportion for $n=1,000$ is 0.9942.

To determine

Explain the advantage of a larger sample size.

Explanation of Solution

For larger sample size the probability becomes approximately 1.

There is higher chance that the probability would be within $\pm 0.04$ of the population proportion.