A paired difference experiment yielded the following results. $$n _ { d } = 50 , \quad \sum { d } = 967 , \quad \sum d ^ { 2 } = 19,201$$
Test $H _ { 0 } : \mu _ { d } = 20$ against $H _ { \mathrm { a } } : \mu _ { d } \neq 20$, where $\mu _ { d } = \mu _ { 1 } - \mu _ { 2 }$, using $\alpha = .05$.

Question

A paired difference experiment yielded the following results. $$n _ { d } = 50 , \quad \sum  { d } = 967 , \quad \sum d ^ { 2 } = 19,201$$
Test $H _ { 0 } : \mu _ { d } = 20$ against $H _ { \mathrm { a } } : \mu _ { d } \neq 20$, where $\mu _ { d } = \mu _ { 1 } - \mu _ { 2 }$, using $\alpha = .05$.

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The Answer of A paired difference experiment yielded the following results. $$n _ { d } = 50 , \quad \sum  { d } = 967 , \quad \sum d ^ { 2 } = 19,201$$
Test $H _ { 0 } : \mu _ { d } = 20$ against $H _ { \mathrm { a } } : \mu _ { d } \neq 20$, where $\mu _ { d } = \mu _ { 1 } - \mu _ { 2 }$, using $\alpha = .05$.

A Paired Difference Experiment Yielded the Following Results nd=50,∑d=967,∑d2=19,201n _ { d } = 50 , \quad \sum { d } = 967 , \quad \sum d ^ { 2 } = 19,201nd​=50,∑d=967,∑d2=19,201

A Paired Difference Experiment Yielded the Following Results $n _ { d } = 50 , \quad \sum { d } = 967 , \quad \sum d ^ { 2 } = 19,201$