Answer:
0.025 = 2.5% probability that the sample proportion will differ from the population proportion by more than 0.03
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
. So
If 314 are sampled, what is the probability that the sample proportion will differ from the population proportion by more than 0.03?
Either it differs by 0.03 or less, or it differs by more than 0.03. The sum of these probabilities is decimal 1.
Probability it differs by 0.03 or less
pvalue of Z when X = 314*(0.06 + 0.03) = 28.26 subtracted by the pvalue of Z when X = 314*(0.06 - 0.03) = 9.42
X = 28.26
has a pvalue of 0.9875
X = 9.42
has a pvalue of 0.0125
0.9875 - 0.0125 = 0.9750
More than 0.03
p + 0.975 = 1
p = 0.025
0.025 = 2.5% probability that the sample proportion will differ from the population proportion by more than 0.03