Question

A random sample of } 260 \text { workers in a high-rise }\end{array}\( office building revealed that \)30 \%\( were very satisfied with the quality of elevator service. At the \)99 \%$ level, what is your estimate of the population value?

Short answer

22.71% to 37.29%

Step-by-step solution

- Identify the sample proportion

The sample proportion is the percentage of the sample that is very satisfied with the elevator service. The given sample proportion is 30%, which can be written as a decimal: \[ \text{Sample Proportion} (p) = 0.30 \]

- Determine the sample size (n)

The sample size, which is the number of workers surveyed, is given as 260: \[ n = 260 \]

- Find the Z-score for 99% confidence level

For a 99% confidence level, the Z-score is approximately 2.576.

- Calculate the standard error

The standard error (SE) can be calculated using the formula: \[ SE = \sqrt{ \frac{p(1-p)}{n} } \] Substituting the values, we get: \[ SE = \sqrt{ \frac{0.30(1-0.30)}{260} } \approx 0.0283 \]

- Calculate the Margin of Error (ME)

The margin of error can be calculated using the formula: \[ ME = Z \times SE \] Substituting the values, we get: \[ ME = 2.576 \times 0.0283 \approx 0.0729 \]

- Compute the confidence interval

Finally, the confidence interval is calculated by adding and subtracting the margin of error from the sample proportion: \[ \text{Lower Limit} = p - ME = 0.30 - 0.0729 \approx 0.2271 \] \[ \text{Upper Limit} = p + ME = 0.30 + 0.0729 \approx 0.3729 \]

- Conclusion

The 99% confidence interval for the population proportion that is very satisfied with the quality of elevator service is approximately from 22.71% to 37.29%.

Key concepts

Sample Proportion

To understand the concept of the sample proportion, let's start by defining it. The sample proportion is a measure that tells us the fraction or percentage of individuals in a sample who have a certain characteristic. In our exercise, we surveyed 260 workers, and 30% of them reported being very satisfied with the elevator service.
In mathematical terms, we write the sample proportion, denoted as \(p\), as:
\[ p = \frac{x}{n} \] Here, \(x\) is the number of satisfied workers (30% of 260) and \(n\) is the total sample size (260). Converting 30% to a decimal, we have: \[ p = 0.30 \]
This sample proportion serves as our best estimate of the population proportion, assuming our sample is representative.

Z-score

The Z-score is a critical value we use in statistics to understand how far away a certain point is from the mean, in terms of standard deviations. When we talk about confidence intervals, the Z-score helps us determine the range where we believe the true population parameter lies, given a confidence level.
For our exercise, we wanted a 99% confidence level. The corresponding Z-score for 99% confidence is approximately 2.576, which we can find using standard Z-tables or statistical software.
\[ Z = 2.576 \]
Using this Z-score in our calculations tells us that if we were to take numerous samples, 99% of the confidence intervals we calculate would contain the true population proportion.

Standard Error

The standard error (SE) is a measure that indicates the variability of our sample proportion. It helps us understand how much our sample proportion is expected to fluctuate from the true population proportion. Calculating the standard error is crucial for determining the precision of our estimate.
We use the formula:
\[ SE = \sqrt{ \frac{p(1-p)}{n}} \]
In this exercise, our sample proportion \(p\) is 0.30 and our sample size \(n\) is 260. Plugging in these values, we get:
\[ SE = \sqrt{ \frac{0.30(1-0.30)}{260} } \approx 0.0283 \]
The standard error thus provides an understanding of how much we can expect the sampling proportion to vary from the real population proportion.

Margin of Error

The margin of error (ME) represents the range of values below and above the sample proportion within which the actual population parameter is expected to lie, given our level of confidence. It combines the variability shown by the standard error with our desired confidence level measured by the Z-score.
Using the formula:
\[ ME = Z \times SE \]
With our Z-score of 2.576 and our standard error of approximately 0.0283:
\[ ME = 2.576 \times 0.0283 \approx 0.0729 \]
This margin of error means that we expect the true population proportion to be within this range (plus or minus from our sample proportion) with 99% confidence.

Confidence Level

The confidence level is the degree of certainty we have that our confidence interval contains the true population parameter. A higher confidence level implies a wider margin of error, reflecting greater certainty, and vice versa.
In our exercise, we chose a 99% confidence level, indicating we want to be 99% sure that the true population proportion of workers satisfied with the elevator service falls within our calculated interval.
The confidence interval in this exercise was calculated as:
\[ 0.30 \pm 0.0729 \]
Which gives us a range from approximately 0.2271 to 0.3729, or 22.71% to 37.29%.
This range means we're 99% confident that the true satisfaction proportion lies within this interval.