A Toyota dealer wants to find out the age of their customers (for advertising purposes). They want the margin of error to be 3 years old. Suppose the sigma is 13 years old. If they want a 90% confidence interval, how many people do they need to know about?
90% = mean ± 1.645 SEm
SEm = SD/√n
3 = 1.645 SEm
at least 47
To determine the sample size needed, we can use the formula for calculating the sample size for a population mean with a specified margin of error and confidence level. The formula is given by:
n = (Z * σ / E)^2
Where:
n = sample size
Z = z-score corresponding to the desired confidence level
σ = standard deviation of the population
E = margin of error
In this case, the desired confidence level is 90%, which corresponds to a z-score of 1.645 (obtained from a standard normal distribution table). The margin of error is 3 years, and the population standard deviation (σ) is given as 13 years.
Substituting the values into the formula:
n = (1.645 * 13 / 3)^2
n = 7.45063290115882^2
n ≈ 55.5154445
Since we cannot have a fraction of a person, we would need to round up the sample size. Therefore, the Toyota dealer would need approximately 56 people to determine the age of their customers with a 90% confidence level and a margin of error of 3 years.