Chapter 9 - The T- and F- Distributions

CH 9: The t- and F- Distributions [t = \frac{\bar{X} - \mu}{S/\sqrt{n}}] From chapter 8, Z = (X̄ - μ) / σ has the standard normal distribution. But when σ, the single "true" value for the standard deviation in the population, is replaced by S, the estimate of σ, this is no longer true, although the approximation is good for large samples. For small samples, it is far from the truth. This is because the sample variance s² (also S²) is itself a random variable. We know that the size of the sample influences the accuracy of our estimates. The larger the sample, the closer the estimate is likely to be to the true value.   The shape of the t-distribution is similar to that of the normal distribution. However, the shape of the t-distribution varies with the sample size. It is longer-tailed (heavier-tailed) than the normal distribution when the sample size is small. As the sample size increases, the t-distribution and normal distribution become progressively closer, and ultimately, they are identical.   Degrees of Freedom Remember, S² = (1/(n-1)) * Σ(x_i - x̄)² To achieve a given sample mean for n numbers, five of those can be chosen at will, but then the last is fully determined. In general, we are told the mean of n numbers is X̄, and that the first n-1 numbers are x₁, x₂, ..., x_(n-1). Image Analysis: t-Distribution and Degrees of Freedom Textual Content: We say that S², based on a sample size n, has n-1 degrees of freedom. Degrees of freedom: For each parameter we need to estimate prior to evaluating the current parameter of interest, we lose one degree of freedom. Because the shape of the t-distribution varies with the sample size, there is a whole family of t-distributions. It is therefore necessary to have some means of indicating which t-distribution is being used in a specific situation. We do this by using a subscript. Intuition suggests that the subscript should be the sample size, but it turns out that the sensible subscript is the degrees of freedom of the standard deviation, which is one less than the sample size.   t = (X̄ - μ) / (S/√n) Exact p-values are usually computed using statistical software. For the z-test, we could calculate these exactly since our tables gave us probabilities associated with z scores. But due to the fact that our t-tables report critical values rather than probabilities, p-values can no longer be computed exactly. To obtain approximate p-values: Identify the appropriate degrees of freedom. Highlight the relevant line (corresponding to the df) in t-tables.   Identify where the test statistic would lie along that line. Determine the approximate probability (range) associated with the test statistic by looking at the probabilities corresponding with the two values on either side of the test statistic.   Confidence Interval for μ When σ² is Estimated by S² If we have a random sample of size n from a population with a normal distribution, and the sample mean is x̄ and the sample variance is s², [(\overline{x} - t_{n-1}\cdot\frac{s}{\sqrt{n}},\overline{x} + t_{n-1}\cdot\frac{s}{\sqrt{n}})] where the t* values are obtained from the t-tables. For 95% confidence intervals, use the column headed 0.025. For 99% confidence intervals, use the column headed 0.005. Testing whether the mean is a specified value (population variance estimated from the sample) Example: A poultry farmer is investigating ways of improving profitability of his operation. Using a standard diet, turkeys grow to a mean mass of 4.5 kg at age 4 months. A sample of 20 turkeys, given a special enriched diet, had an average mass of 4.8 kg after 4 months. The sample standard deviation was 0.5 kg. Using a 5% significance level, test whether the new diet is effectively increasing mass.   H₀: μ ≤ 4.5 (one-sided alternative: we assume enriched diet should not cause loss of mass) H₁: μ > 4.5 Significance level: 5% Rejection region: Because the population variance is unknown, we need to use the t-distribution. Degrees of freedom for t will be 19 (based on 20 observations). We will reject H₀ if the observed t-value exceeds t₀.₀₅(19) Test statistic: t = (4.8 - 4.5) / (0.5 / √20) = 2.68 Conclusion: 2.68 > 1.729 We reject H₀ and conclude that, at the 5% significance level, we have established that the new enriched diet is effective. Comparing two sample means: Matched pairs or paired samples Example: Consider a medicine designed to reduce dizziness. We want to know if it is effective. We have 10 patients who complain of dizziness and we examine the reduction in dizzy spells for each patient after taking medicine. [How many dizzy spells had per month before the meds and again while on meds]   Before (B): 19 18 9 7 12 16 22 19 18 10 After (A): 17 24 12 4 15 19 25 16 24 6 d = B - A: 2 -6 -3 4 0 -3 -3 3 -6 The two samples have been effectively reduced to one by taking the difference in scores. Therefore, testing if the two populations ("before" and "after") have the same mean is equivalent to testing if the (population) mean of the difference scores is 0. So, the two paired samples have been reduced to one sample on which we can perform a one-sample t-test. The relevant summary statistics obtained from the data are d̄ = -1.5 and sd = 3.57. Performing the hypothesis test assuming a 5% level of significance:   H₀: μ_d = 0 H₁: μ_d > 0 α = 0.05 Critical value is 1.833 If the number of dizzy spells have reduced while on the treatment, the before score should be greater than the after score. And since difference were defined by d = B - A, we'd expect them to be positive. Note that the sample info is not consistent with this hypothesis. Hypotheses are determined by what you are interested in finding out, NOT by the data!   Once again we assume that this is a random representative sample from the population of differences and that the differences in reading speed are normally distributed with mean μ_d.   H₀: μ_d = 0 H₁: μ_d ≠ 0 t = -1.77 The p-value for this statistic lies between 0.1 and 0.2. Therefore we do not reject the null hypothesis (p-value is too large) and conclude that there isn't a difference in reading speed between the two methods.   Confidence Intervals As before, it is also possible to construct confidence intervals for these differences. Following the same logic as before and using the fact that d̄ ~ tn-1, the 95% confidence interval can be constructed by:   Pr(-t_(n-1,0.025) ≤ d̄ ≤ t_(n-1,0.025)) = 0.95 So Pr(d̄ - t_(n-1,0.025) * sd/√n ≤ μ_d ≤ d̄ + t_(n-1,0.025) * sd/√n) = 0.95 Therefore, the 95% confidence interval is: (-4.015, 1.515)