The Food Marketing Institute shows that 18% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.18 and a sample of 900 households will be selected from the population
What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,200 households (to 4 decimals)?
To find the probability that the sample proportion will be within +/- 0.02 of the population proportion, we can use the concept of the sampling distribution of the sample proportion.
The sampling distribution of the sample proportion follows an approximate normal distribution when the sample size is sufficiently large, which is the case here with 1,200 households.
The standard deviation of the sampling distribution of the sample proportion can be calculated using the formula:
Standard Deviation = sqrt((p * (1-p)) / n)
Where:
- p is the population proportion (0.18)
- n is the sample size (1,200)
Substituting the values into the formula, we get:
Standard Deviation = sqrt((0.18 * (1-0.18)) / 1,200)
Simplifying the equation, we get:
Standard Deviation ≈ sqrt(0.0001482) ≈ 0.01216
Since we want to find the probability of the sample proportion being within +/- 0.02 of the population proportion, we need to find the probability of the sample proportion falling within the range of (p - 0.02, p + 0.02).
Using the standard normal distribution table or a statistical calculator, we can find the probability of the sample proportion falling within this range. The z-scores corresponding to p - 0.02 and p + 0.02 can be calculated as:
z1 = (p - 0.02 - p) / Standard Deviation = -0.02 / 0.01216 ≈ -1.64
z2 = (p + 0.02 - p) / Standard Deviation = 0.02 / 0.01216 ≈ 1.64
Finding the probability associated with these z-scores will give us the probability of the sample proportion falling within the desired range.
Using a standard normal distribution table or a statistical calculator, we find that the cumulative probability for a z-score of -1.64 is approximately 0.0495. Similarly, the cumulative probability for a z-score of 1.64 is approximately 0.9505.
To find the probability of the sample proportion falling within the desired range, we subtract the lower cumulative probability from the higher cumulative probability:
Probability = 0.9505 - 0.0495 ≈ 0.901
Therefore, the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,200 households is approximately 0.901, or 90.1% (to 4 decimals).