People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior.
a)What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion (to 4 decimals)?
b)What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion (to 4 decimals)?
a. µ = np = 540(.12) = 64.8
σ = √(.12*.88/540) = 0.014
z = 0.03/0.014 = 2.14
z = -0.03/0.014 = -2.14
P(-2.14 < z < 2.14) =0.9676
b. µ = np = 540(.12) = 64.8
σ = √(.12*.88/540) = 0.014
z = 0.015/0.014 = 1.07
z = -0.015/0.014 = -1.07
P(-1.07 < z < 1.07) = 0.7154
Thanks for your help I really appreciate it.
You're welcome
To find the probability in this case, we will be using the sampling distribution of the sample proportion. The formula for the standard deviation of the sample proportion is given by:
σ = sqrt(p * (1 - p) / n)
Where:
- σ is the standard deviation of the sample proportion
- p is the population proportion (in this case, 0.12)
- n is the sample size (in this case, 540)
a) For the first question, we need to find the probability that the sample proportion falls within ±0.03 of the population proportion. To do this, we need to calculate the z-scores for both boundaries and use the standard normal distribution.
The z-score is given by:
z = (x - p) / σ
Where:
- z is the z-score
- x is the value we want to find the probability for (±0.03)
- p is the population proportion (0.12)
- σ is the standard deviation of the sample proportion, which we previously calculated.
Let's calculate the z-scores for both boundaries:
z1 = (0.12 - 0.03) / σ
z2 = (0.12 + 0.03) / σ
Then, we can use a standard normal distribution table or a calculator to find the probability associated with each z-score. To find the probability within the two boundaries, we subtract the probability associated with z1 from the probability associated with z2:
P = P(z1 ≤ Z ≤ z2) = P(z2) - P(z1)
b) For the second question, the process is the same. We need to find the probability that the sample proportion falls within ±0.015 of the population proportion. Again, we calculate the z-scores for both boundaries:
z1 = (0.12 - 0.015) / σ
z2 = (0.12 + 0.015) / σ
Then, we use a standard normal distribution table or a calculator to find the probability associated with each z-score and subtract the probability associated with z1 from the probability associated with z2:
P = P(z1 ≤ Z ≤ z2) = P(z2) - P(z1)
Remember to round the probabilities to 4 decimals.