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Suppose a level .05 test of \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} > 0\) is to be performed, assuming \((m = n)\)\({\sigma _1} = {\sigma _2} = \)10 and normality of both distributions, using equal sample sizes. Evaluate the probability of a type II error when \({\mu _1} - {\mu _2} = 1\) and\(n = 25,100,2500\), and 10,000. Can you think of real problems in which the difference \({\mu _1} - {\mu _2} = 1\)has little practical significance? Would sample sizes of n=10,000 be desirable in such problems?

Short Answer

Expert verified

For\(n = 25 - \beta = 0.9015\);

For\(n = 100 - \beta = 0.8264\);

For\(n = 2500 - \beta = 0.0294\);

For\(n = 10,000 - \beta = 0\).

Step by step solution

01

To evaluate the probability of a type II error probability

Type II Error Probability \(\beta \) for a Level \(\alpha \)Test, when alternative hypothesis is\({H_a}:{\mu _1} - {\mu _2} > {\Delta _0}\), and \(n = m\) is

\(\beta = \Phi \;\;\;{z_\alpha } + \frac{{{\Delta _0}}}{{\sigma /\sqrt n }}\)

Thus, the type II error probability for n=25 and \({z_{0.05}} = 1.645\) is

\({\beta _1} = \Phi \;\;\;1.645 + \frac{1}{{14.142/\sqrt {25} }} = 0.9015.\)

The type II error probability for \(n = 100\)and \({z_{0.05}} = 1.645\) is

\({\beta _2} = \Phi \;\;\;1.645 + \frac{1}{{14.142/\sqrt {100} }} = 0.8264\)

The type II error probability for \(n = 25\) and \({z_{0.05}} = 1.645\) is

\({\beta _3} = \Phi \;\;\;1.645 + \frac{1}{{14.142/\sqrt {2500} }} = 0.0294\)

The type II error probability for \(n = 25\)and \({z_{0.05}} = 1.645\) is

\({\beta _4} = \Phi \;\;\;1.645 + \frac{1}{{14.142/\sqrt {10,000} }} = 0\)

\(\begin{array}{l}{\rm{All values were computed using software}}{\rm{. However, }}\\{\rm{you could use the table in the appendix}}{\rm{. }}\end{array}\)

02

Final proof 

For\(n = 25 - \beta = 0.9015\);

For\(n = 100 - \beta = 0.8264\);

For\(n = 2500 - \beta = 0.0294\);

For\(n = 10,000 - \beta = 0\).

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Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 3 19.70 1.10 0.65

2 3 10.90 0.60 0.35

Estimate for difference: 8.800

95% CI for difference: 6.498,11.102

T -Test of difference =0 (vs not = ):

T -Value =12.1 P -Value =0.001 DF=3

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