Question

In a recent poll of 1100 randomly selected home delivery truck drivers, 26% said they had encountered an aggressive dog on the job at least once. Report the 95% confidence interval for the proportion of all home delivery truck drivers who have encountered an aggressive dog on the job at least once. (Round final calculations to the nearest tenth of a percent)

a) 24.5% to 27.5%
b) 25.8% to 26.2%
c) 20.3% to 31.7%
d) 26.9% to 27.1%

Answer 1

The 95% confidence interval for the proportion of home delivery truck drivers who have encountered an aggressive dog on the job is between 24.5% and 27.5%, calculated using the standard formula for a confidence interval for proportions.

Step-by-step explanation:

To calculate the 95% confidence interval for the proportion of home delivery truck drivers who have encountered an aggressive dog on the job at least once, we use the formula for a confidence interval for a proportion:

Confidence Interval = p ± z*sqrt[p(1-p)/n]

Where:

• p = sample proportion
• z = z-score corresponding to the desired confidence level
• n = sample size

Using the information given:

1. p = 0.26 (26% encountered an aggressive dog)
2. n = 1100 (number of participants in the poll)

The z-score for a 95% confidence level is approximately 1.96. Plugging the numbers into the formula, we compute the confidence interval:

Confidence Interval = 0.26 ± 1.96*sqrt[0.26*0.74/1100]

After calculating, we round the final calculations to the nearest tenth of a percent to obtain the interval. The correct interval that includes the sample proportion within the bounds provided is option a) 24.5% to 27.5%. Thus, we can say with 95% confidence that the true proportion of all home delivery truck drivers who have encountered an aggressive dog on the job at least once lies between 24.5% and 27.5%.