Question

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

Short answer

Without an exact t-distribution table, precise \(P\) -values can't be found. However, for lower \(P\) -values (below 0.05), the null hypothesis would generally be rejected. A t score of 1.6 (given n=16), 3.2 (given n=14) and 6.3 (given n=16) would likely have a significant \(P\) -value in a two-tailed test, leading to rejection of the null hypothesis. A t score of -5.1 (given n=20) in a one-tailed test would also likely be significant because it is in the left tail of the t distribution.

Step-by-step solution

Identify degrees of freedom and locate t scores for each scenario

The degree of freedom (\(df\)) here equals \(n-1\) for each situation. Now, for each situation, locate the corresponding score on a t-distribution table.

Calculate the P-value for each scenario

For each situation, the \(P\) -value represents the probability that the observed data (or data more extreme) occurred by chance given the null hypothesis is true. Use the degrees of freedom (\(df\)) and the t scores to find the relative \(P\) -values from a t-distribution table. For two-tailed tests, this value has to be multiplied by 2.

Interpret the P-value in context of the test

The \(P\) -value corresponds to the level of significance. If the \(P\) -value falls below the predetermined alpha level (usually 0.05), the null hypothesis is rejected. Inspecting the \(P\) -values from each case will give a sense of whether the null hypothesis should be rejected or not. Conclusions must be drawn based on these \(P\) -values. If there isn't a t-distribution table available, it may be commented that the \(P\) -value cannot be calculated exactly without one.