What is the probability that a simple random sample of 56 special orders will
provide an estimate of the population proportion of special orders specifying the color black that is within plus or minus .05 of the actual population proportion,assuming p = .36? In other words, what is the probability that p will be between .31
and .41?
To find the probability that the sample proportion will be within plus or minus 0.05 of the actual population proportion, we can use the concept of sampling distribution and the normal approximation to the binomial distribution.
1. Calculate the standard error of the sample proportion (SE):
SE = sqrt(p * (1 - p) / n)
where p is the actual population proportion and n is the sample size.
In this case, p = 0.36 and n = 56, so the SE is:
SE = sqrt(0.36 * (1 - 0.36) / 56) = 0.0608
2. Convert the boundaries of interest (.31 and .41) into z-scores:
z1 = (0.31 - p) / SE
z2 = (0.41 - p) / SE
Using the z-table or a statistical calculator, find the corresponding probabilities for the z-scores. The probability we're interested in is the difference between these two probabilities.
Please note that this calculation assumes a large enough sample size (n > 30) for the normal approximation to be valid. If the sample size is small, it would be more appropriate to use the exact binomial distribution.
Alternatively, you can use statistical software like R, Python, or Excel to calculate this probability directly using functions like "pnorm" or "norm.dist".