Problem

Use the Student's $t$-distribution to find the $t$-value for each of the given scenarios. Round $t$-values to four decimal places.
- Find the value of $t$ such that the area in the left tail of the $t$-distribution is 0.025 , if the sample size is 57 .
\[
t=
\]
- Find the value of $t$ such that the area in the right tail of the $t$-distribution is 0.05 , if the sample size is 74 .
\[
t=
\]
- Find the value of $t$ such that the area in the right tail of the $t$-distribution is 0.01 , if the sample size is 60 .
\[
t=
\]
- Find the two values of $t$ such that $95 \%$ of the area under the $t$-distribution is centered around the mean, if the sample size is 16 . Enter the solutions using a comma-separated list.
\[
t=
\]

Answer

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Answer

\(\boxed{t = -2.1314, 2.1314}\)

Steps

Step 1 :Calculate the degrees of freedom (df) by subtracting 1 from the sample size. For the first problem, df = 57 - 1 = 56.

Step 2 :Look up the $t$-value in the $t$-distribution table or use a $t$-distribution calculator with df = 56 and left tail area = 0.025. The $t$-value is approximately -2.0047.

Step 3 :\(\boxed{t = -2.0047}\)

Step 4 :Calculate the degrees of freedom (df) for the second problem. df = 74 - 1 = 73.

Step 5 :Look up the $t$-value in the $t$-distribution table or use a $t$-distribution calculator with df = 73 and right tail area = 0.05. The $t$-value is approximately 1.6672.

Step 6 :\(\boxed{t = 1.6672}\)

Step 7 :Calculate the degrees of freedom (df) for the third problem. df = 60 - 1 = 59.

Step 8 :Look up the $t$-value in the $t$-distribution table or use a $t$-distribution calculator with df = 59 and right tail area = 0.01. The $t$-value is approximately 2.3901.

Step 9 :\(\boxed{t = 2.3901}\)

Step 10 :Calculate the degrees of freedom (df) for the fourth problem. df = 16 - 1 = 15.

Step 11 :Look up the $t$-values in the $t$-distribution table or use a $t$-distribution calculator with df = 15 and the area in the two tails = 1 - 0.95 = 0.05. The $t$-values are approximately -2.1314 and 2.1314.

Step 12 :\(\boxed{t = -2.1314, 2.1314}\)

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