These ten exercises follow the companion guide on sampling distributions. They run from the mean and variance of the sample mean, through required sample size and the Central Limit Theorem, to the chi-squared, t, and F distributions and the art of recognising which one a combination of normals produces. Work each one fully before reading the solutions, which explain the reasoning rather than just stating the answer.
Exercises
Exercise 1. A population has mean 50 and variance 144. A random sample of size 36 is drawn. Find the expected value, the variance, and the standard deviation of the sample mean, and state whether the population must be normal for these results to hold.
Exercise 2. Weights are normally distributed with mean 70 and variance 25. A random sample of size 25 is taken. Find the probability that the sample mean exceeds 72. Use the fact that the probability a standard normal exceeds 2 is 0.0228.
Exercise 3. A normal population has standard deviation 10. Find the smallest sample size n for which the sample mean lies within 2 units of the true mean with probability at least 0.95. Use the critical value 1.96.
Exercise 4. Claim processing times are heavily right-skewed with mean 4 minutes and variance 16. A random sample of size 100 is taken. Approximate the probability that the sample mean exceeds 4.5 minutes, and justify the method you use. The probability a standard normal exceeds 1.25 is 0.1056.
Exercise 5. Let six independent standard normal variables be squared and summed. State the distribution of the sum, and give its mean and variance. Then, if one chi-squared variable has 6 degrees of freedom and an independent one has 9, state the distribution of their sum.
Exercise 6. A random sample of size 10 is drawn from a normal population with variance 20. Using that nine times the sample variance divided by 20 follows a chi-squared distribution with 9 degrees of freedom, and that the probability this exceeds 16.92 is 0.05, find the value c for which the probability that the sample variance exceeds c is 0.05.
Exercise 7. Let a standard normal variable and an independent chi-squared variable with 8 degrees of freedom be combined into a t variable with 8 degrees of freedom. Given that the probability it exceeds 1.86 is 0.05 and the probability it exceeds 2.31 is 0.025, bound the probability that it exceeds 2.0, and find the probability that it falls below minus 1.86.
Exercise 8. An independent pair of chi-squared variables, one with 4 degrees of freedom and one with 8, is combined into an F variable with 4 and 8 degrees of freedom. Given that the probability an F variable with 8 and 4 degrees of freedom exceeds 6.04 is 0.05, find the value below which the original F variable falls with probability 0.05.
Exercise 9. Three independent normal variables each have mean 0 and variance 9. The probability a standard normal exceeds 1.15 is 0.1251, and the probability a chi-squared variable with 2 degrees of freedom exceeds 5.99 is 0.05.
Exercise 10. Three independent normal variables each have mean 0 and variance 4. The probability an F variable with 1 and 2 degrees of freedom exceeds 18.5 is 0.05.
Solutions
Solution 1. The sample mean is centred on the population mean, so its expected value is 50. Its variance is the population variance divided by the sample size.
These results require only that the observations are independent and identically distributed with finite mean and variance. The population does not need to be normal.
Solution 2. Because the population is normal, the sample mean is exactly normal, centred at 70 with variance equal to 25 divided by 25, which is 1, so its standard deviation is 1.
Standardising the threshold gives a z value of 2.
Solution 3. The requirement is that the sample mean lies within 2 of the mean with probability 0.95, which after standardising becomes a condition on the half-width in z units.
Solving for the square root of n gives at least 9.8, so n must be at least 96.04. Since the sample size is a whole number, round up.
Solution 4. The population is far from normal, but the sample of 100 is large, so the Central Limit Theorem justifies treating the sample mean as approximately normal. Its mean is 4 and its variance is 16 divided by 100, which is 0.16, giving a standard deviation of 0.4.
Standardising the threshold gives a z value of 1.25.
Solution 5. A sum of squared independent standard normals is chi-squared with degrees of freedom equal to the number of terms, here 6.
Its mean equals the degrees of freedom and its variance is twice that, so the mean is 6 and the variance is 12. By the additivity of independent chi-squared variables, the degrees of freedom add, so the sum of the two given variables has 15 degrees of freedom.
Solution 6. The scaled sample variance follows a chi-squared distribution with 9 degrees of freedom.
The upper 5 percent point of that distribution is 16.92, so the event that the scaled variance exceeds 16.92 has probability 0.05. Translating back to the sample variance gives the threshold.
So the value sought is 37.6, meaning the probability the sample variance exceeds 37.6 is 0.05.
Solution 7. The combination is a t variable with 8 degrees of freedom. The value 2.0 lies between the listed points 1.86 and 2.31, whose upper-tail probabilities are 0.05 and 0.025, so the probability of exceeding 2.0 lies between them.
By the symmetry of the t distribution about zero, the lower tail mirrors the upper tail.
Solution 8. The lower 5 percent point of an F distribution is found from the upper point of the reciprocal distribution with its degrees of freedom swapped.
So the F variable with 4 and 8 degrees of freedom falls below 0.166 with probability 0.05.
Solution 9. For part (a), a linear combination of independent normals is normal, with the variances adding after squaring the coefficients. Here each variance is 9 and there are three terms with coefficients of magnitude 1, so the total variance is 27 and the standard deviation is the square root of 27, about 5.196.
Standardising the threshold of 6 gives a z value of about 1.15.
For part (b), dividing each variable by 3 gives a standard normal, so each squared and scaled term is chi-squared with one degree of freedom, and the sum of two is chi-squared with two.
The upper 5 percent point of that distribution is 5.99, so translating back gives the threshold.
So the probability that the sum of the two squares exceeds 53.91 is 0.05.
Solution 10. For part (a), the difference of the three independent normals is itself normal, centred at zero with variance 4 plus 4 plus 4, which is 12.
Because the distribution is symmetric about zero, the probability of exceeding zero is exactly one half.
For part (b), dividing each variable by 2 gives a standard normal, so each squared and scaled term is chi-squared with one degree of freedom. Forming the ratio of the first squared term to the sum of the other two, each divided by its degrees of freedom, gives an F variable with 1 and 2 degrees of freedom.
The stated event that the first square exceeds 9.25 times the sum of the other two is exactly the event that this F ratio exceeds 18.5.
See you soon.
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