The proportion of all consumers favoring a new product might be a slow as 0.20 or as high as 0.60 a random sample is to be used to estimate the proportion of the consumers who favor the new product to with in plas or - 0.05 with a confidence cofficient of 90% to be on the safe ( large sample) size what sample size should be used?
To calculate the required sample size, we need to use the formula:
n = (z^2 * p * q) / e^2
where:
z = the z-score corresponding to the desired confidence level (90% = 1.645)
p = the proportion of consumers favoring the new product (we'll use the midpoint of the given range, p = 0.40)
q = 1 - p
e = the maximum error margin allowed (0.05)
Plugging in the values, we get:
n = (1.645^2 * 0.4 * 0.6) / 0.05^2
n = 141.384
Rounding up to the nearest whole number, we get a required sample size of 142. Therefore, we would need to survey at least 142 consumers to estimate the proportion favoring the new product within plus or minus 0.05 with 90% confidence.
To determine the sample size needed to estimate a proportion with a specific level of confidence, we can use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-score for the desired level of confidence (in this case, 90%, which corresponds to a Z-score of 1.645)
p = estimated proportion of consumers favoring the new product
E = margin of error
Given that the estimated proportion of consumers favoring the new product can vary from 0.20 to 0.60, we can use the worst-case scenario (p = 0.20) to make sure we have a larger sample size. The margin of error is 0.05.
Substituting the values into the formula:
n = (1.645^2 * 0.20 * (1 - 0.20)) / 0.05^2
n = (2.706 * 0.20 * 0.80) / 0.0025
n = 0.43248 / 0.0025
n ≈ 172.992
Therefore, a sample size of approximately 173 should be used to estimate the proportion of consumers favoring the new product with a margin of error of ±0.05 and a 90% confidence level.