Step 1 :The problem is asking for the t-value such that the area to the right of it under the t-distribution curve with 6 degrees of freedom is 0.15. This is equivalent to finding the 85th percentile (1 - 0.15 = 0.85) of the t-distribution with 6 degrees of freedom.
Step 2 :We can use the Percent Point Function (PPF), which is the inverse of the Cumulative Distribution Function (CDF), to find this value. The PPF gives the variable value for a given percentile of a distribution.
Step 3 :Let's denote the degrees of freedom as \(df = 6\) and the percentile as \(p = 0.85\).
Step 4 :By using the PPF, we find that the t-value is approximately 1.134.
Step 5 :So, the value of \(t_{0.15}\) is \(\boxed{1.134}\).