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The confidence level of the three intervals can be computed as

1 - left tail area - right tail area,

which can be said to be the central area. On the pictures below you can see the central area for all three intervals.

(a): reminder: the area to the left of \({\rm{ - 0}}{\rm{.687 is 0}}{\rm{.25}}\) (given in the exercise) and the area to the right of \({\rm{1}}{\rm{.725 is 0}}{\rm{.05}}\),

Therefore, the central area is\(1 - 0.25 - 0.05 = 0.7\)or \(70\% \)confidence level.

(b): reminder: the area to the left of \({\rm{ - 0}}{\rm{.860 is 0}}{\rm{.2}}\) (given in the exercise) and the area to the right of \({\rm{1}}{\rm{.325 is 0}}{\rm{.0}}{\rm{.1}}\),

Therefore, the central area is\(1 - 0.2 - 0.1 = 0.7\)or \(70\% \)confidence level.

(c): reminder: the area to the left of \({\rm{ - 1}}{\rm{.064 is 0}}{\rm{.15}}\) (given in the exercise) and the area to the right of \({\rm{1}}{\rm{.064 is 0}}{\rm{.15}}\), therefore, the central area is

\(1 - 0.15 - 0.15 = 0.7\)or \(70\% \)confidence level.

The confidence level for all three intervals is the same.

The width of the interval shall decide which interval to recommend. The width can be computed by subtracting the lower bound from the upper bound.

(a): for this interval, the width is

\({\rm{\bar x + 0}}{\rm{.687}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}{\rm{ - }}\left( {{\rm{\bar x - 1}}{\rm{.725}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}} \right){\rm{ = 2}}{\rm{.412 \times }}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}\)

(b): for this interval, the width is

\({\rm{\bar x + 0}}{\rm{.860}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}{\rm{ - }}\left( {{\rm{\bar x - 1}}{\rm{.325}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}} \right){\rm{ = 2}}{\rm{.185 \times }}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}\)

(c): for this interval, the width is

\({\rm{\bar x + 1}}{\rm{.064}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}{\rm{ - }}\left( {{\rm{\bar x - 1}}{\rm{.064}}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}} \right){\rm{ = 2}}{\rm{.128 \times }}\frac{{\rm{s}}}{{\sqrt {{\rm{21}}} }}\)

Obviously, the width of the third interval is the smallest, and the interval is symmetrical, therefore it should be used.

For a t interval the symmetric interval is always the best choice.

Hence The third interval, its width is the shortest and it is symmetric.