Q1. What is the probability that a randomly selected American adult has a Life Satisfaction score within 30 points of the population mean (i.e., between 470 and 530)?
The population distribution is normal with a mean of 500 and a standard deviation of 100.
First we calculate the standardized z score corresponding to 530 to be z = (x – m)/s = (530 – 500)/100 = 30/100 = 0.3, and the z score corresponding to 470 is (470-500)/100 = -30/100 = -.30. Then we use thez table, or the p-z calculator, to find that the probability that a randomly selected score has a z value greater than 0.3 is 0.382, and the probability that this score has a z value less than -.3 is also 0.382. Thus, the probability of observing a z score between -0.3 and +.3 is 1.000 – .(382 + .382) = 1.000 – .764 = .236, or about 24%.
Q2. What is the probability that a randomly selected sample of N=100 American adults has a mean Life Satisfaction score within 30 points of the population mean?
If we knew the standard deviation of the sampling distribution of the mean, we could calculate the zscore for any observed mean and then use a z table, or the p-z calculator, to find the probability of observing a mean within any specific range.
Fortunately, from the Central Limit Theorem we know that the standard error of the mean (i.e., the standard deviation of the sampling distribution of the mean) is equal to the standard deviation of the population divided by the square root of the sample size. In our example where the standard deviation of the population is 100 and the sample size is also 100, the standard error of the mean is:
Thus, a sample mean that is 10 points greater than the population mean has a z score of 1.00.
In our example, we would like to know how likely it is that a sample mean differs from the population mean by 30 points. This corresponds to a z score of 30/10 = 3. The probability of observing a sample mean between 470 and 530 in our example is the same as the probability of observing a z score between -3.0 and +3.0 on a standardized normal distribution. From the z table, or the p-z calculator, we find the probability of observing a z score greater than 3.0 is only 0.001; the probability of observing a zscore less than -3.0 is also 0.001. Thus, the probability of observing a z score between -3.0 and +3.0 is 1.0000 – (0.001 + 0.001) = 0.998.
We conclude that the probability is over 99% that a randomly selected sample of N=100 American adults will have a mean Life Satisfaction score within 30 points of the actual population mean.
Q3. What is the probability that a randomly selected sample of N=25 American adults has a mean Life Satisfaction score within 30 points of the population mean?
Using the Central Limit Theorem, we can find the standard error of the mean by dividing the standard deviation of the population by the square root of the sample size:
A sample mean that deviates 30 points from the population mean corresponds to a z score of 30/20 = 1.5. From the z table, or the p-z calculator, we find the probability of observing a z score greater than 1.5 is 0.067; the probability of observing a z score less than -1.5 is also 0.067. Thus, the probability of observing a z score between -1.5 and 1.5 is 1.000 – (0.067 + 0.067) = 1.000 – 0.134 = 0.866.
We conclude that the probability is about 87% that a randomly selected sample of N= 25 American adults will have a mean Life Satisfaction score within 30 points of the actual population mean.
Q4. What is the probability that a randomly selected sample of N=5 American adults has a mean Life Satisfaction score within 30 points of the population mean?
The standard error of the mean is:
A sample mean that is 30 points greater than the population mean has a z score of 30/44.72 = 0.67. From the z table, or the p-z calculator, we find the probability of observing a z score greater than 0.67 is 0.251. Thus, the probability of observing a z score between -0.67 and +0.67 is 1.0000 – (0.251 + 0.251) = 1.000 – 0.502 = 0.498.
We conclude that the probability is about 50% that a randomly selected sample of N=5 American adults will have a mean Life Satisfaction score within 30 points of the actual population mean.
Q5. Complete the following table and comment on the relationship between sample size and the expected accuracy of a sample mean:
| Sample size | Probability that a sample mean differs from the population mean by more than 30 points |
| 100 | 0.2%* |
| 25 | 13% |
| 5 | 50% |
*In Q2 we determined that there is a 99.8% chance that the mean of a sample of N=100 will be within 30 points of the population mean. Thus, the probability that a mean for a sample withN=100 will differ from the population mean by more than 30 points is 100% – 99.8%, or 0.2%. You can use your findings in Q3 and Q4 to calculate the values for samples of N=25 and N=5.
We notice from the table that as the sample size increases, the probability decreases that our sample mean differs from the population mean by more than 30 points. Thus, larger samples are more likely to give us accurate estimates of the population mean. The mean of a small sample, such as N=5, is relatively unstable and much more likely to differ substantially from the population mean.
