A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 560 college students showed that 27% of them had, or intended to, cheat on examinations. Find the 95% confidence interval.
A. 0.2323 to 0.3075
B. 0.2325 to 0.3075
C. 0.2325 to 0.3185
D. 0.2323 to 0.3185
Need answer
To find the 95% confidence interval for the proportion of college students who cheat on exams, we can use the formula:
Confidence interval = Sample proportion ± Margin of error
First, let's calculate the sample proportion. In this case, the sample proportion is 27% or 0.27 (since it's given that 27% of the 560 college students admitted to cheating).
Next, let's calculate the margin of error. The margin of error depends on the sample size and the desired level of confidence. For a 95% confidence level, the standard margin of error can be calculated using the formula:
Margin of error = Z * (sqrt((p * (1-p)) / n))
Where:
- Z is the z-value for the desired confidence level (for a 95% confidence level, Z = 1.96).
- p is the sample proportion.
- n is the sample size.
Substituting the given values into the formula, we have:
Margin of error = 1.96 * (sqrt((0.27 * (1-0.27)) / 560))
Calculating this expression, we find that the margin of error is approximately 0.03127 (rounded to five decimal places).
Now we can calculate the confidence interval. Plugging in the values into the formula for the confidence interval, we have:
Confidence interval = 0.27 ± 0.03127
Calculating the lower and upper bounds of the confidence interval, we find:
Lower bound = 0.27 - 0.03127 = 0.23873 (rounded to five decimal places)
Upper bound = 0.27 + 0.03127 = 0.30127 (rounded to five decimal places)
Therefore, the 95% confidence interval is 0.23873 to 0.30127.
Looking at the answer choices provided, we can see that none of them match the calculated confidence interval exactly. However, the closest option is:
C. 0.2325 to 0.3185
So, the closest answer choice is C.