Null and alternate hypotheses

Null and alternate hypotheses           

 

Week X Y XY XX 1 29 21 609 841 2 35 32 1120 1225 3 41 11 451 1681 4 32 26 832 1024 Total ∑X = 137 ∑Y = 90 ∑ XY = 3012 ∑XX = 4771 Given n = 4, ∑X = 137 Mean = ∑X / n Standard Deviation (s) = √ ∑{(X – x)(X – x)} / n Mean = 137 / 4 Standard Deviation (s) = √ 17689 / 4 Mean = 34.25 Standard Deviation (s) = √ 4422.25 Standard Deviation (s) = 66.5 The null and alternate hypothesis here becomes:

H0: μ ≥ 34.25

H1: μ < 34.25

The hypothesis test statistics is:

Z = (x-μ) / (σ-√n)

= (34.25-35) / (0.05-√100)

= -0.75 / -9.95

= 0.075

Probability (p) value when (σ) = 0.05, and its meaning

p-value = P(Z>0.075) = 0.0001

The probability of getting a sample whose mean is 34.25 or more when H0 is true is 0.0001.

Comparing the p – value to alpha

For a large test statistic, the critical value is large and its p-value is small than even the alpha value.A rejection is made if test statistics is found in the rejection region.