Angela is conducting a poll on campus to determine the next student representative. How many students does she need to sample for a confidence level of 90% with a margin of error of + or - 5%?
degree of freedom for a chi-test depends on the sample size true or false
To Ana:
Try this formula:
n = [(z-value)^2 * p * q]/E^2
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 .05 (5%) in the problem. Z-value is found using a z-table (for 90%, the value is 1.645).
Note also that ^2 means squared and * means to multiply.
I'll let you take it from here. Be sure to round the value to the next highest whole number.
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To Anonymous:
Degrees of freedom for a chi-square test depends on the number of rows and columns.
To calculate the sample size needed for a poll with a given confidence level and margin of error, we can use the formula:
n = (Z^2 * p * (1-p)) / e^2
Where:
- n is the sample size,
- Z is the Z-score corresponding to the desired confidence level,
- p is the estimated proportion of the population (unknown unless you have data),
- e is the margin of error.
In this case, we have a confidence level of 90%, meaning the desired Z-score is 1.645. The margin of error is ±5%, which can be written as 0.05. Since we don't have an estimated proportion (p) of the population, we can use the conservative assumption of 0.5, which gives the maximum required sample size.
Plugging these values into the formula:
n = (1.645^2 * 0.5 * (1-0.5)) / 0.05^2
After calculating that expression, the required sample size will be obtained.