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If you select a very small value for αwhen conducting a hypothesis test, will β tend to be big or small? Explain.

Short Answer

Expert verified

The lower values of significance level (α) increase the type II error (β).

Step by step solution

01

Given information

The information regarding the values αand β.

The level of significance, αis small.

02

Define significance level and type II error probability

Type II error probability (β):

The type II error probability βis calculated assuming that the null hypothesis is false because it is defined as the probability of accepting H0when it is false.

i.e; β=P(acceptH0/H0isfalse0

Significance level (α):

The level of significance is the size of the type I error. In other words, rejecting the null hypothesis when it is true.

i.e.α=P(rejectH0/H0istrue)

03

Explain about  β when small value of  α conducting a hypothesis test

The type II error probabilityaccepts the null hypothesis when it is false. So, if a significance level is very small, then the region of acceptance expands. As a result, makes it difficult to reject the null hypothesis.

Therefore the lower values of significance level increase the type II error.

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Most popular questions from this chapter

Solder-joint inspections. Current technology uses high-resolution X-rays and lasers to inspect solder-joint defects on printed circuit boards (PCBs) (Global SMT & Packaging, April 2008). A particular manufacturer of laser-based inspection equipment claims that its product can inspect, on average, at least 10 solder joints per second when the joints are spaced .1 inch apart. The equipment was tested by a potential buyer on 48 different PCBs. In each case, the equipment was operated for exactly 1 second. The number of solder joints inspected on each run follows:

The potential buyer wants to know whether the sample data refute the manufacturer’s claim. Specify the null and alternative hypotheses that the buyer should test.

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Play Golf America program. The Professional Golf Association (PGA) and Golf Digest have developed the Play Golf America program, in which teaching professionals at participating golf clubs provide a free 10-minute lesson to new customers. According to Golf Digest, golf facilities that participate in the program gain, on average, \(2,400 in greens fees, lessons, or equipment expenditures. A teaching professional at a golf club believes that the average gain in greens fees, lessons, or equipment expenditures for participating golf facilities exceeds \)2,400.

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