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a. Show for the upper-tailed test with \({\sigma _1}\) and \({\sigma _2}\)known that as either\(m\) or\(n\) increases, \(\beta \)decreases when \({\mu _1} - {\mu _2} > {\Delta _0}\).

b. For the case of equal sample sizes \(\left( {m = n} \right)\)and fixed \(\alpha \),what happens to the necessary sample size \(n\) as \(\beta \) is decreased, where \(\beta \) is the desired type II error probability at a fixed alternative?

Short Answer

Expert verified

the solution is

The corresponding \({z_\beta }\) increases as the intended \(\beta \) (type II error) lowers. When the sample sizes are equivalent, \({z_\beta }\) is in the numerator of the formula for necessary sample size \(n\), indicating that when \({z_\beta }\) rises, the sample size increases as well (see exercise 13).

Step by step solution

01

show the upper-tailed test

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)represents the null hypothesis. The type II error \(\beta \) for \({\mu _1} - {\mu _2} = {\Delta ^'}\) vary depending on the alternative hypothesis. The alternative hypothesis is \({H_a}:{\mu _1} - {\mu _2} > 0\), which indicates that the type II error is present

\(\beta \left( {{\Delta ^'}} \right) = \Phi \left( {{z_\alpha } - \frac{{\Delta ' - {\Delta _0}}}{\sigma }} \right)\)

Where

\(\sigma = \sqrt {\frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}} \)

When \(n\) or \(m\) goes up, \(\sigma \) goes down (because we divide with bigger number). This implies that.

\(\frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }\)

Because the numerator is positive, the value will rise. As a result, the number.

\({z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }\)

Because \(\Phi \) is a cdf of standard normal distribution, the type Il error lowers.

\(\Phi \left( {{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }} \right)\)

Will decreases as well.

02

type II error probability

The corresponding \({z_\beta }\) increases as the intended \(\beta \) (type II error) lowers. When the sample sizes are equivalent, \({z_\beta }\) is in the numerator of the formula for necessary sample size \(n\), indicating that when \({z_\beta }\) rises, the sample size increases as well (see exercise 13).

03

conclusion

The corresponding \({z_\beta }\) increases as the intended \(\beta \) (type II error) lowers. When the sample sizes are equivalent, \({z_\beta }\) is in the numerator of the formula for necessary sample size \(n\), indicating that when \({z_\beta }\) rises, the sample size increases as well (see exercise 13).

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Most popular questions from this chapter

The article "Fatigue Testing of Condoms" cited in Exercise 7.32 reported that for a sample of 20 natural latex condoms of a certain type, the sample mean and sample standard deviation of the number of cycles to break were 4358 and 2218 , respectively, whereas a sample of 20 polyisoprene condoms gave a sample mean and sample standard deviation of 5805 and 3990 , respectively. Is there strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles? Carry out a test using a significance level of 01 . (Note: The cited paper reported P-values of t tests for comparing means of the various types considered.)

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Before

After

S

F

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\({{\bf{X}}_{\bf{1}}}\)

\({{\bf{X}}_{\bf{2}}}\)

F

\({{\bf{X}}_{\bf{3}}}\)

\({{\bf{X}}_{\bf{4}}}\)

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\(T = \frac{{(\bar X - \bar Y) - \left( {{\mu _1} - {\mu _2}} \right)}}{{{S_p}\sqrt {\frac{1}{m} + \frac{1}{n}} }}\)

which has a\(t\)distribution with\(m + n - 2\)df when both population distributions are normal with\({\sigma _1} = {\sigma _2}\)(see the Pooled\(t\)Procedures subsection for a description of\({S_p}\)).

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