In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.
a) How many randomly selected employers must we contact in order to create an estimate in which we are confident with a margin of error of ?
b) Suppose we want to reduce the margin of error to . What sample size will suffice?
c) Why might it not be worth the effort to try to get an interval with a margin of error of ?
a) A sample size of is needed.
(Round up to the nearest whole number.)

Ask by Mcguire Davies. in the United States

Mar 28,2025

Question

In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.
a) How many randomly selected employers must we contact in order to create an estimate in which we are confident with a margin of error of ?
b) Suppose we want to reduce the margin of error to . What sample size will suffice?
c) Why might it not be worth the effort to try to get an interval with a margin of error of ?
a) A sample size of is needed.
(Round up to the nearest whole number.)

Ask by Mcguire Davies. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

Let $$ p $$ be the true proportion of employers planning to hire additional employees. The margin of error for estimating a proportion is given by

$$
E = z_{\alpha/2} \sqrt{\frac{p(1-p)}{n}}
$$

In a conservative approach we use $$ p = 0.5 $$ since this maximizes $$ p(1-p) $$. Thus, the formula for the required sample size becomes

$$
n = \frac{z_{\alpha/2}^2 \cdot 0.25}{E^2}
$$

### Part (a)

For a margin of error $$ E = 0.07 $$ (or $$ 7\% $$) and a $$ 99\% $$ confidence level, we have

$$
z_{\alpha/2} = 2.576
$$

Substitute into the formula:

$$
n = \frac{(2.576)^2 \cdot 0.25}{(0.07)^2}
$$

Calculate $$ (2.576)^2 $$:

$$
(2.576)^2 \approx 6.635
$$

Thus,

$$
n = \frac{6.635 \cdot 0.25}{0.0049} = \frac{1.6588}{0.0049} \approx 338.33
$$

Rounding up to the nearest whole number:

$$
n = 339
$$

### Part (b)

For a margin of error $$ E = 0.03 $$ (or $$ 3\% $$) and still a $$ 99\% $$ confidence level, use

$$
n = \frac{(2.576)^2 \cdot 0.25}{(0.03)^2}
$$

Calculate $$ (0.03)^2 $$:

$$
(0.03)^2 = 0.0009
$$

Thus,

$$
n = \frac{6.635 \cdot 0.25}{0.0009} = \frac{1.6588}{0.0009} \approx 1843.11
$$

Rounding up to the nearest whole number:

$$
n = 1844
$$

### Part (c)

Reducing the margin of error to $$ 1\% $$ (or $$ E = 0.01 $$) results in the sample size

$$
n = \frac{(2.576)^2 \cdot 0.25}{(0.01)^2} = \frac{6.635 \cdot 0.25}{0.0001} = \frac{1.6588}{0.0001} \approx 16588
$$

A sample size of approximately $$ 16,\!588 $$ is very large. This may not be worth the effort because:

1. **Cost:** Sampling such a large number of employers can be very expensive.
2. **Time:** Collecting data from so many sources may require a lot of time.
3. **Logistical Constraints:** Managing and processing that many responses can be impractical.
4. **Diminishing Returns:** The increase in precision (from a $$ 7\% $$ to a $$ 1\% $$ margin of error) may not justify the additional effort and resources needed.

### Final Answers

- (a) A sample size of $$ \boxed{339} $$ is needed.
- (b) A sample size of $$ \boxed{1844} $$ is needed.
- (c) It might not be worth the effort to obtain a $$ 1\% $$ margin of error because the required sample size (approximately $$ 16,\!588 $$) is very large, making it cost-prohibitive and logistically challenging.
- (a) 339 employers are needed. - (b) 1844 employers are needed. - (c) It's not worth the effort to get a 1% margin of error because the sample size would be too large and impractical.

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