Final answer:
The 80% confidence interval for the proportion of students at the college who have completed their required English 101 course is approximately 0.418 to 0.614.
Step-by-step explanation:
To construct an 80% confidence interval for the proportion of students who have completed their required English 101 course, we will use the sample proportion to estimate the population proportion and then add and subtract the margin of error to this sample proportion.
The sample proportion (π) is the number of students who said 'yes' (48) divided by the total number of students surveyed (93), so π = 48/93 = 0.516
Next, we'll use the z-score for an 80% confidence level, which is approximately 1.282. The formula for the margin of error (ME) is:
ME = z * √ (π(1-π) / n), where n is the sample size.
Lower Bound ≈ 0.516 - 1.282 × √(0.516×0.484)/93
Upper Bound ≈ 0.516 + 1.282 × √(0.516×0.484)/93
After calculating the margin of error, we'll add and subtract it from the sample proportion to find the lower and upper limits of the confidence interval.
Lower Bound≈0.418
Upper Bound≈0.614
Therefore, the 80% confidence interval for the proportion (p) of students who have completed their required English 101 course is:
0.418<p<0.614