Workers in several industries were surveyed to determine the proportion of workers who feel their industry is understaffed. In the government sector, 37% of the respondents said they were understaffed, in the health care sector 33% said they were understaffed, and in the education sector 28% said they were understaffed (USA Today, January 11, 2010). Suppose that 200 workers were surveyed in each industry.

a. Construct a 95% confidence interval for the proportion of workers in each of these industries who feel their industry is understaffed (to 4 decimals). Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use commas in your numerical answer (i.e. use 1200 instead of 1,200, etc.)
Government industry
Health care industry
Education industry

b. Assuming the sample size will be the same in each industry, how large would the sample need to be to ensure that the margin of error is .05 or less for each of the three confidence intervals? (Round sample size (n) up to the next integer value.)

a. To construct a confidence interval for the proportion of workers who feel their industry is understaffed, we can use the formula for a confidence interval for a proportion:

Confidence Interval = Sample Proportion ± Margin of Error

The sample proportion is the proportion of workers who feel their industry is understaffed, and the margin of error determines the range in which the true proportion is likely to fall.

1) Government industry:
Sample Proportion (p1) = 37% = 0.37
Sample Size (n1) = 200

Margin of Error (E1) = z * sqrt((p1 * (1 - p1)) / n1), where z is the critical value from the standard normal distribution corresponding to the desired level of confidence. For a 95% confidence interval, z is approximately 1.96.

E1 = 1.96 * sqrt((0.37 * (1 - 0.37)) / 200) ≈ 0.0422

Confidence Interval = 0.37 ± 0.0422 ≈ (0.3278, 0.4122)

2) Health care industry:
Sample Proportion (p2) = 33% = 0.33
Sample Size (n2) = 200

Margin of Error (E2) = z * sqrt((p2 * (1 - p2)) / n2)

E2 = 1.96 * sqrt((0.33 * (1 - 0.33)) / 200) ≈ 0.0403

Confidence Interval = 0.33 ± 0.0403 ≈ (0.2897, 0.3703)

3) Education industry:
Sample Proportion (p3) = 28% = 0.28
Sample Size (n3) = 200

Margin of Error (E3) = z * sqrt((p3 * (1 - p3)) / n3)

E3 = 1.96 * sqrt((0.28 * (1 - 0.28)) / 200) ≈ 0.0385

Confidence Interval = 0.28 ± 0.0385 ≈ (0.2415, 0.3185)

Confidence intervals for each industry:
Government industry: (0.3278, 0.4122)
Health care industry: (0.2897, 0.3703)
Education industry: (0.2415, 0.3185)

b. The formula for margin of error in the confidence interval for proportions is:

Margin of Error (E) = z * sqrt((p * (1 - p)) / n)

We want the margin of error to be 0.05 or less for each confidence interval.

To find the required sample size (n) to achieve this, we can rearrange the formula for margin of error:

n = (z^2 * p * (1 - p)) / E^2

Using the largest value of p (0.37), and the maximum margin of error (0.05), we can calculate the required sample size:

n = (1.96^2 * 0.37 * (1 - 0.37)) / 0.05^2 ≈ 335.8856

Rounding up to the next integer, the sample size (n) should be at least 336 for each industry to ensure a margin of error of 0.05 or less for each confidence interval.

a. To construct a confidence interval for the proportion of workers in each industry who feel their industry is understaffed, we can use the formula for a confidence interval for a proportion:

Confidence Interval = Sample Proportion ± Margin of Error

The sample proportion is the proportion of workers who responded "understaffed" in each industry. The margin of error depends on the sample size and the level of confidence.

For the government industry:
Sample Proportion = 37% = 0.37
Margin of Error = z * sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size)
Since the survey was conducted on 200 workers in each industry, we have:
Sample Size = 200
Level of Confidence = 95%
Therefore, z = 1.96 (from the Z-table for a 95% confidence level)

Plugging in the values:
Margin of Error = 1.96 * sqrt((0.37 * (1 - 0.37)) / 200) = 1.96 * sqrt(0.2356 / 200) = 0.0678

Confidence Interval = Sample Proportion ± Margin of Error = 0.37 ± 0.0678 = (0.3022, 0.4378)

Similarly, for the health care industry:
Sample Proportion = 33% = 0.33
Margin of Error = 1.96 * sqrt((0.33 * (1 - 0.33)) / 200) = 0.0676

Confidence Interval = 0.33 ± 0.0676 = (0.2624, 0.3976)

And for the education industry:
Sample Proportion = 28% = 0.28
Margin of Error = 1.96 * sqrt((0.28 * (1 - 0.28)) / 200) = 0.0675

Confidence Interval = 0.28 ± 0.0675 = (0.2125, 0.3475)

Therefore, the 95% confidence intervals for the proportion of workers in each industry who feel their industry is understaffed are as follows:

Government industry: (0.3022, 0.4378)
Health care industry: (0.2624, 0.3976)
Education industry: (0.2125, 0.3475)

b. To determine the sample size needed to ensure that the margin of error is 0.05 or less for each of the three confidence intervals, we need to rearrange the formula for the margin of error:

Margin of Error = z * sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size)

We want the margin of error to be 0.05 or less, so we have:

0.05 ≤ z * sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size)

Simplifying the equation:

sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size) ≤ 0.05 / z

We can square both sides of the inequality to eliminate the square root:

(Sample Proportion * (1 - Sample Proportion)) / Sample Size ≤ (0.05 / z)^2

Since we want the worst-case scenario, we assume Sample Proportion = 0.5, which gives us the maximum value of the fraction.

(0.5 * (1 - 0.5)) / Sample Size ≤ (0.05 / z)^2

0.25 / Sample Size ≤ (0.05 / z)^2

Sample Size ≥ 0.25 / (0.05 / z)^2

Using the z-values for a desired confidence level of 95%, we know that z = 1.96.

Sample Size ≥ 0.25 / (0.05 / 1.96)^2

Sample Size ≥ 0.25 / (0.0024256)

Sample Size ≥ 103.0927

To ensure that the margin of error is 0.05 or less for each of the three confidence intervals, the sample size would need to be at least 104 (rounding up to the next integer) for each industry.

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