the average age for licensed drivers in the country is u=40.3 years with a standard deviation of o= 13.2 years.
a. a researcher obtained a random sample of n=16 parking tickets and computed an average age of m=38.9 years for the drivers. is it reasonable to conclude that this set of n=16 people is a representative sample of licensed drivers?
b. the same researcher obtained a random sample of n=36 speeding tickets and computed an average age of m=36.2 years for the drivers. compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. is it reasonable to conclude that this set of n= 36 people is a representative sample of licensed drivers?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
a. To determine whether the sample of n=16 parking ticket holders is a representative sample, we can perform a hypothesis test using the z-test. The null hypothesis (H0) would be that the sample is representative, and the alternative hypothesis (Ha) would be that the sample is not representative.
The formula to calculate the z-score is:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Given:
Population mean (µ) = 40.3 years
Standard deviation (σ) = 13.2 years
Sample mean (x̄) = 38.9 years
Sample size (n) = 16
Calculating the z-score:
z = (38.9 - 40.3) / (13.2 / sqrt(16))
z = -1.4 / (13.2 / 4)
z = -1.4 / 3.3
z ≈ -0.424
To determine if it is reasonable to conclude that the sample is representative, we need to compare the calculated z-score with a critical value, usually denoted as α (alpha). The critical value depends on the desired level of significance. Let's assume a common significance level of 0.05 (or 5%).
Using a z-table or a statistical software, we can find that the critical value for a two-tailed test at a 5% significance level is approximately ±1.96.
Since |z| < 1.96, we fail to reject the null hypothesis and conclude that the sample of n=16 parking ticket holders is a reasonable representative sample of licensed drivers.
b. To calculate the z-score for the sample mean and find the probability, we will follow similar steps as in part (a).
Given:
Sample mean (x̄) = 36.2 years
Sample size (n) = 36
Calculating the z-score:
z = (36.2 - 40.3) / (13.2 / sqrt(36))
z = -4.1 / (13.2 / 6)
z = -4.1 / 2.2
z ≈ -1.864
Now, we want to find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Since we have a left-tailed question (we want to find the probability of being younger than the sample mean), we need to find the corresponding area under the standard normal distribution curve.
Using a z-table or a statistical software, we can find that the probability for a z-score of -1.864 is approximately 0.0322, or about 3.22%.
To determine if it is reasonable to conclude that the sample is representative, we need to compare the probability to the level of significance. With a probability of 3.22%, which is less than the commonly used significance level of 5%, we have strong evidence to reject the null hypothesis. Therefore, we conclude that the sample of n=36 speeding ticket holders is not a representative sample of licensed drivers.