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

Question

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

Examlex · Accepted Answer

The Answer is

s _ { d } ^ { 2 } = \frac { 19,201 - 967 ^ { 2 } / 50 } { 49 } \approx 10.188 , ext { so } s _ { d } \approx 3.19

The test statistic is

2 = \frac { 19.34 - 20 } { 3.19 / \sqrt { 50 } } \approx - .1 .46

The rejection region is

z < - 1.96

or

z > 1.96

.

Since the test statistic does not fall in the rejection region, we have insufficient evidence to conclude that the mean difference is not 20 . difference is not 20.

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

Definitions:

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

Definitions:

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