QUESTION IMAGE
Question
determine the t-value in each of the cases
click the icon to view the table of areas under the t-distribution
(a) find the t-value such that the area in the right tail is 0.25 with 5 degrees of freedom
0.727 (round to three decimal places as needed.)
(b) find the t-value such that the area in the right tail is 0.15 with 13 degrees of freedom
1.079 (round to three decimal places as needed.)
(c) find the t-value such that the area left of the t-value is 0.025 with 19 degrees of freedom. hint: use symmetry
(round to three decimal places as needed.)
Step1: Recall the property of the t - distribution symmetry
The t - distribution is symmetric about \(t = 0\). That is, if \(P(T\lt t_1)=\alpha\), then \(P(T\gt - t_1)=\alpha\) for a t - random variable \(T\) with \(n\) degrees of freedom.
We are given that the area to the left of the t - value is \(0.025\) with \(n = 19\) degrees of freedom. Let the t - value be \(t\). So \(P(T\lt t)=0.025\). Using the symmetry property of the t - distribution \(P(T\gt - t)=0.025\).
Step2: Use the t - distribution table
We look up the value in the t - distribution table. For a right - tail area of \(\alpha=0.025\) and \(n = 19\) degrees of freedom.
In the t - distribution table, we find the row corresponding to \(n = 19\) degrees of freedom and the column corresponding to the right - tail area. The value in the table for a right - tail area of \(0.025\) and \(19\) degrees of freedom is \(t = 2.093\)
Since \(P(T\lt t)=0.025\) and by symmetry \(t=- 2.093\)
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\(-2.093\)