Question

A small company involved in e-commerce is interested in statistics concerning online purchases. A poll found that 42% of a random sample of 1,187 adults, said they don't buy products or services online.

(a) Find the margin of error, if we want 95% confidence in our estimate of the percent of American adults who don't buy products or services online: (Use 4 decimals.)


(b) Find the margin of error for 99% confidence: (Use 4 decimals.)

Answer 1

The margin of error at 95% confidence is 0.0282, and at 99% confidence, it is 0.0367. These margins of error help to understand the range within which the true proportion of American adults not purchasing online likely falls.

Step-by-step explanation:

To find the margin of error for a proportion with a given confidence level, we use the formula:

Margin of Error (E) = Z * sqrt[(p*(1-p))/n]

where:

• Z is the Z-score corresponding to the desired confidence level
• p is the sample proportion
• n is the sample size

For part (a), using a 95% confidence level, the Z-score is approximately 1.96. For part (b), a 99% confidence level corresponds to a Z-score of approximately 2.576. The sample proportion (p) is 0.42, and the sample size (n) is 1,187.

Calculations:

1. 95% Confidence Level:
   
   E = 1.96 * sqrt[(0.42*(1-0.42))/1187] = 0.0282 (rounded to four decimal places)
2. 99% Confidence Level:
   
   E = 2.576 * sqrt[(0.42*(1-0.42))/1187] = 0.0367 (rounded to four decimal places)

The margin of error gives us a range within which we can be confident that the true proportion lies. It is important to note that while the margin of error accounts for statistical sampling variability, it does not account for all potential forms of survey error, such as non-response bias or measurement errors.