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.
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?
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