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Chapter 8: Q10E (page 325)

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special

laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

Short Answer

Expert verified

The relevant hypotheses are \({H_0}:{\mu _1} = {\mu _2}\)versus \({H_a}:{\mu _1} > {\mu _2}\).

Step by step solution

01

Errors in Hypothesis testing.

A type I error consists of rejecting the null hypothesis H0 when it is true.

A type II error involves not rejecting H0 when it is false.

02

Step 2:Test statistic.

A test statistic is a function of the sample data used as a basis for deciding whether H0 should be rejected. The selected test statistic should discriminate effectively between the two hypotheses. That is, values of the statistic that tend to result when H0 is true should be quite different from those typically observed when H0 is not true.

03

Hypothesis results.

Before stating the relevant hypotheses, denote with \({\mu _1}\) the average for the regular laminate, and with \({\mu _2}\) the average for the special laminate.

The relevant hypotheses are \({H_0}:{\mu _1} = {\mu _2}\)versus \({H_a}:{\mu _1} > {\mu _2}\).

The type I error is to conclude that the war page for special laminate is less than the regular laminate when it is not.

The type II error is to conclude that there is no difference in the laminates when the special laminate produces less war page.

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Most popular questions from this chapter

Verify that the piecewise-defined function \(y = \left\{ {\begin{array}{*{20}{r}}{ - {x^2},}&{x < 0}\\{{x^2},}&{x \ge 0}\end{array}} \right.\) is a solution of the differential equation \(xy' - 2y = 0\) on \(( - \infty ,\infty )\).

The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\overline x = 94.32\). Assume that the distribution of the melting point is normal with 蟽 =1.20.

a.Test H0: 碌 =95 versus Ha: 碌鈮� 95 using a two -tailed level .01 test.

b.If a level .01 test is used, what is 尾(94), the probability of a type II error when 碌=94?

c.What value of n is necessary to ensure that 尾(94) = .1 when 伪 = .01?

Reconsider the accompanying sample data on expense ratio (%) for large-cap growth mutual funds first introduced in Exercise 1.53.

\(\begin{array}{l}0.52 1.06 1.26 2.17 1.55 0.99 1.10 1.07 1.81 2.05\\0.91 0.79 1.39 0.62 1.52 1.02 1.10 1.78 1.01 1.15\end{array}\)

A normal probability plot shows a reasonably linear pattern.

a. Is there compelling evidence for concluding that the population mean expense ratio exceeds \(1\% \)? Carry out a test of the relevant hypotheses using a significance level of \(.01\).

b. Referring back to (a), describe in context type I and II errors and say which error you might have made in reaching your conclusion. The source from which the data was obtained reported that \(\mu = 1.33\) for the population of all \(762\) such funds. So, did you actually commit an error in reaching your conclusion?

c. Supposing that \(\sigma = .5\), determine and interpret the power of the test in (a) for the actual value of m stated in (b).

For the following pairs of assertions, indicate which with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples):

a. H0: 碌= 100, Ha: 碌 > 100

b.H0: 蟽= 20, Ha: \(\sigma \le 20\)

c.H0: p鈮� .25, Ha: p= .25

d.H0: 碌1 - 碌2 = 25, Ha: 碌1 - 碌2 > 100

e.H0: \(S_1^2 = S_2^2\) , Ha: \(S_1^2 \ne S_2^2\)

f.H0: 碌= 120, Ha: 碌= 150

g.H0: 蟽1,/蟽2 =1,Ha: 蟽1,/ 蟽2 鈮�1

h.H0p1 鈥� p2 = -.1, Ha: p1 鈥� p2 < -.1

Let 碌 denote the true average radioactivity level (picocuries per liter). The value 5 pCi/L is considered the dividing line between safe and unsafe water. Would you recommend testing H0: 碌= 5 versus Ha: 碌> 5 or H0: 碌= 5 versus Ha: 碌 < 5? Explain your reasoning. (Hint: Think about the consequences of a type I and type II error for each possibility.)
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