Final answer:
To calculate a 98% confidence interval for the proportion of students with part-time jobs based on a sample of 26 out of 100, a specific confidence interval formula involving the sample proportion and a Z-score for the 98% level is used. Without the provided formulas (a or b) being shown, it is impossible to confirm which is correct, leading to the response of 'None of the above'.
This correct answer is C.
Step-by-step explanation:
The question is asking for the correct formula to calculate a 98% confidence interval for the proportion of all college students with part-time jobs, given that 26 out of 100 sample students have part-time jobs. The standard formula for a confidence interval for a population proportion starts with the sample proportion (p), which in this case is 26/100 or 0.26. The formula can be written as:
p ± Z*sqrt((p(1-p))/n)
Where ± denotes 'plus or minus', Z is the Z-score corresponding to the desired confidence level (which can be found using a Z-table or standard statistical software for a 98% confidence level), sqrt represents the square root function, p is the sample proportion, and n is the sample size.
To answer the student's question, neither Formula a nor Formula b was provided, thus we cannot determine whether one of them is correct.
Without the actual formulas, we would conclude with 'None of the above' if neither of the provided formulas matches the standard formula for a confidence interval for a population proportion. To calculate the actual confidence interval, you need to substitute the known values into the formula and solve.
This correct answer is C.