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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

Automatic identification of the boundaries of significant structures within a medical image is an area of ongoing research. The paper 鈥淎utomatic Segmentation of Medical Images Using Image Registration: Diagnostic and

Simulation Applications鈥� (J. of Medical Engr. and Tech., 2005: 53鈥�63)discussed a new technique for such identification. A measure of the accuracy of the automatic region is the average linear displacement (ALD). The paper gave the following ALD observations for a sample of 49 kidneys (units of pixel dimensions).

1.38 0.44 1.09 0.75 0.66 1.28 0.51

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1.30 0.57 0.43 0.62 1.00 1.05 0.82

1.10 0.65 0.99 0.56 0.56 0.64 0.45

0.82 1.06 0.41 0.58 0.66 0.54 0.83

0.59 0.51 1.04 0.85 0.45 0.52 0.58

1.11 0.34 1.25 0.38 1.44 1.28 0.51

a.Summarize/describe the data.

b.Is it plausible that ALD is at least approximately normally distributed? Must normality be assumed prior to calculating a CI for true average ALD or testing hypotheses about true average ALD? Explain.

c.The authors commented that in most cases the ALD is better than or of the order of 1.0. Does the data in fact provide strong evidence for concluding that true average ALD under these circumstances is less than 1.0? Carry out an appropriate test of hypotheses.

d.Calculate an upper confidence bound for true average ALD using a confidence level of 95%, and interpret this bound.

Newly purchased tires of a particular type are supposed to be filled to a pressure of 30 psi. Let 碌 denote the true average pressure. A test is to be carried out to decide whether 碌 differs from the target value. Determine the P-value for each of the following z test statistic values.

a.2.10 b. -1.75 c. -.55 d. 1.41 e. -5.3

Each of a group of \(20\) intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let \(p\) denote the proportion of all such players who would prefer gut to nylon, and let \(X\) be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most \(50\% \) of all such players prefer gut. We simplify this to \({H_0}:p = .5\), planning to reject \({H_0}\) only if sample evidence strongly favors gut strings.

a. Is a significance level of exactly \(.05\) achievable? If not, what is the largest a smaller than \(.05\) that is achievable?

b. If \(60\% \) of all enthusiasts prefer gut, calculate the probability of a type II error using the significance level from part (a). Repeat if 80% of all enthusiasts prefer gut.

c. If \(13\) out of the \(20\) players prefer gut, should \({H_0}\) be rejected using the significance level of (a)?

The true average diameter of ball bearings of a certain type is supposed to be .5 in. A one-sample t test will be carried out to see whether this is the case. What conclusion is appropriate in each of the following situations?

a.n= 13, t= 1.6, 伪= .05

b.n= 13, t= -1.6, 伪 =.05

c.n= 25, t= -2.6, 伪 =.01

d.n= 25, t= -3.9

Reconsider the paint-drying situation of Example 8.5, in which drying time for a test specimen is normally distributed with 蟽 = 9. The hypotheses H0: 碌 =75 versus Ha: 碌 <75 are to be tested using a random sample of n= 25 observations.

a.How many standard deviations (of X) below the null value is \(\overline x = 72.3\)?

b.If \(\overline x = 72.3\), what is the conclusion using 伪 =.002?

c.For the test procedure with 伪 =.002, what is 尾(70)?

d.If the test procedure with 伪 =.002 is used, what n is necessary to ensure that 尾(70) = .01?

e.If a level .01 test is used with n5 100, what is the probability of a type I error when m5 76?Answer the following questions for the tire problem in Example 8.7.

a.If \(\overline x = 30,960\) 30,960 and a level 伪=.01 test is used, what is the decision?

b.If a level .01 test is used, what is 尾(30,500)?

c.If a level .01 test is used and it is also required that 尾(30,500) = .05, what sample size n is necessary?

d.If \(\overline x = 30,960\), what is the smallest 伪 at which H0 can be rejected (based on n = 16)?
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