Direct mail advertisers send solicitations (“junk mail”) to thousands of potential customers in the hope that some will buy the company’s product. The
response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1080 people randomly selected from
their mailing list of over 200,000 people. They get orders from 137 of the recipients. Use this information to complete parts a through d.
a) Create a confidence interval for the percentage of people the company contacts who may buy something.
(Round to one decimal place as needed.)

Ask by Best Huff. in the United States

Mar 28,2025

Question

Direct mail advertisers send solicitations (“junk mail”) to thousands of potential customers in the hope that some will buy the company’s product. The
response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1080 people randomly selected from
their mailing list of over 200,000 people. They get orders from 137 of the recipients. Use this information to complete parts a through d.
a) Create a confidence interval for the percentage of people the company contacts who may buy something.
(Round to one decimal place as needed.)

Ask by Best Huff. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

The sample proportion is

$$
\hat{p} = \frac{137}{1080} \approx 0.1269.
$$

The standard error (SE) is given by

$$
\text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.1269 \times 0.8731}{1080}} \approx \sqrt{\frac{0.1108}{1080}} \approx \sqrt{0.0001026} \approx 0.0101.
$$

For a $$90\%$$ confidence interval, the critical value $$z^*$$ is approximately $$1.645$$. The margin of error (ME) is

$$
\text{ME} = z^* \times \text{SE} \approx 1.645 \times 0.0101 \approx 0.0166.
$$

Thus, the $$90\%$$ confidence interval is

$$
\hat{p} \pm \text{ME} \approx 0.1269 \pm 0.0166.
$$

In decimal form, the interval is approximately

$$
(0.1103,\ 0.1435).
$$

Expressed as percentages and rounded to one decimal place, the confidence interval is

$$
(11.0\%,\ 14.4\%).
$$
The $$90\%$$ confidence interval for the percentage of people who may buy something is $$ (11.0\%,\ 14.4\%) $$.

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