I'm stuck on a question in MM207...
Assume you want to estimate with the proportion of students who commute less than 5 miles to work within 2%, what sample size would you need?
could someone please help me ASAP?
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .5 * .5]/.02^2
I'll let you finish.
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .02 (2%) in the problem. Z-value is found using a z-table (for 95%, the value is 1.96).
To estimate the sample size for a given population proportion, we can use the formula:
n = (Z^2 * p * (1 - p)) / E^2
where:
n = required sample size
Z = Z-value corresponding to the desired level of confidence (e.g., 1.96 for a 95% confidence level)
p = estimated proportion (in this case, the proportion of students who commute less than 5 miles to work)
E = margin of error (in this case, 2%)
To determine the required sample size, we need to know the estimated proportion of students who commute less than 5 miles to work. If you have this information or a reasonable estimate for it, we can proceed with calculating the sample size.