Question

A random sample of 700 units from a large consignment showed that 200 were damaged. Find
95% confidence interval for the proportion of damaged unit in the consignment.

Answer 1

Answer on Question #70008 – Math – Statistics and Probability

Question

A random sample of 700 units from a large consignment showed that 200 were damaged. Find 95% confidence interval for the proportion of damaged unit in the consignment.

Solution

First, let's find the sample proportion:

$$\hat {p} = \frac {m}{n} = \frac {200}{700} = \frac {2}{7} \approx 0.285714.$$

Second, let's find the standard error of the sample proportion:

$$\sigma = \sqrt {\frac {\hat {p} (1 - \hat {p})}{n}} = \sqrt {\frac {\frac {2}{7} \left(1 - \frac {2}{7}\right)}{700}} = \sqrt {\frac {\frac {10}{49}}{700}} = \sqrt {\frac {1}{3430}} \approx 0.017075.$$

Error in the sample proportion is believed to follow a normal distribution with $\sigma$ computed above. For a two-tailed confidence interval with the level of confidence of 95%, we need corresponding z-score from a table:

$$z _ {\alpha / 2} = z _ {(1 - 0.95) / 2} = z _ {0.025} = 1.96.$$

Then the margin of error is

$$z _ {0.025} \sigma = 1.96 \times 0.017075 = 0.033467.$$

Finally, the 95% confidence interval is

$$\begin{array}{l} (\hat {p} - z _ {0.025} \sigma , \hat {p} + z _ {0.025} \sigma) = (0.285714 - 0.033467, 0.285714 + 0.033467) \\ = (0.252247, 0.319181). \end{array}$$

Answer: (0.252247, 0.319181).

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