Final answer:
The 99% confidence interval for the proportion of punctuation errors made by beginning programming students is between 0.2205 and 0.2795, calculated using the sample proportion, Z-value for the 99% confidence level, and the given sample size.
Step-by-step explanation:
To find a 99% confidence interval for the proportion of punctuation errors made by beginning programming students, we can use the formula for a confidence interval for a proportion: CI = π ± Z*sqrt[π(1-π)/n], where π is the sample proportion, Z* is the Z-value corresponding to the desired confidence level, and n is the sample size.
First, we calculate the point estimate of the proportion (π) which is the number of punctuation errors divided by the total number of errors, so π = 75/300 = 0.25. Next, we find the corresponding Z-value for the 99% confidence level which is approximately 2.576.
Then, we calculate the margin of error (E) using the Z-value and the sample proportion:
- E = Z* · sqrt[π(1-π)/n]
- E = 2.576 · sqrt[0.25(1-0.25)/300]
- E = 0.0295
Finally, the 99% confidence interval is calculated by subtracting and adding the margin of error from the point estimate:
- Lower Limit = 0.25 - 0.0295 = 0.2205
- Upper Limit = 0.25 + 0.0295 = 0.2795
Therefore, we are 99% confident that the true proportion of punctuation errors among the student programs lies between 0.2205 and 0.2795.