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Refer to Example 9.7. Does the data suggest that the standard deviation of the strength distribution for fused specimens is smaller than that for not-fused specimens? Carry out a test at significance level .01.

Short Answer

Expert verified

Do not reject null hypothesis and variances are equal’

Step by step solution

01

To find the hypothesis

Testing null hypothesis \({H_0}:\sigma _1^2 = \sigma _2^2\)versus alternative hypothesis\({H_a}\), under the assumption that the two populations are normal and independent

Depending on alternative hypothesis \({H_a}\)the \(P\)value is corresponding area under the \({F_{m - 1,n - 1}}\)curve.

The hypotheses of interest are \(H:0:{\sigma _1} = \sigma _2^2\) versus\({H_a}:\sigma _1^2 > \sigma _2^2\), where \({\sigma _1},{\sigma _2}\) denote the variance of the not fused population and fused population, respectively. The test is one sided (upper).

From the exercise 9.7, page 376, the corresponding sample standard deviations are \({s_1} = 277.3\) and\({s_2} = 205.9\). The degrees of freedom are \(m - 1 = 10 - 1 = 9\)and\(n - 1 = 8 - 1 = 7\).

02

Finding test statistic

Thus, the value of the statistic is

\(\begin{array}{c}f = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{277.3}^2}}}{{{{205.9}^2}}}\\ = 1.814\end{array}\)

\(\)

03

Decision rule and conclusion

At significance level\(\alpha = 0.01\), and mentioned degrees of freedom, value \({F_{0.01,9,7}}\) is

\({F_{0.01,9,7}} = 6.72\)

From the table in the appendix (or you could use a software which is better).

The test is upper sided, and

\({F_{0.01,9,7}} = 6.72 > 1.814 = f\)

Hence do not reject null hypothesis at given significance level.

P-value approach:

For the upper sided test, the\(P\)value is

\(\begin{array}{c}P = P(F > 1.814)\\ = 0.2223\end{array}\)

Where \(F\) has Fisher's distribution with degrees of freedom \({\nu _1} = 9\)and \({\nu _2} = 7.\)The value was computed using a software. The \(P\) value is large, and

\(P = 0.2223 > 0.05 = \alpha \)

Thus. do not reject null hypothesis at any reasonable significance level. The variances are equal.

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Most popular questions from this chapter

Using the traditional formula, a \(95\% \) CI for \({p_1} - {p_2}\)is to be constructed based on equal sample sizes from the two populations. For what value of \(n( = m)\)will the resulting interval have a width at most of .1, irrespective of the results of the sampling?

The following summary data on bending strength (lb-in/in) of joints is taken from the article "Bending Strength of Corner Joints Constructed with Injection Molded Splines" (Forest Products J., April, 1997: 89-92).

Type Sample size Sample mean sample SD

Without side coating 10 80.95 9.59

With side coating 10 63.23 5.96

a. Calculate a 95 % lower confidence bound for true average strength of joints with a side coating.

b. Calculate a 95 % lower prediction bound for the strength of a single joint with a side coating.

c. Calculate an interval that, with 95 % confidence, includes the strength values for at least 95 % of the population of all joints with side coatings.

d. Calculate a 95 % confidence interval for the difference between true average strengths for the two types of joints.

An experiment was performed to compare the fracture toughness of high-purity \(18Ni\) maraging steel with commercial-purity steel of the same type (Corrosion Science, 1971: 723–736). For \(m = 32\)specimens, the sample average toughness was \(\overline x = 65.6\) for the high purity steel, whereas for \(n = 38\)specimens of commercial steel \(\overline y = 59.8\). Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial purity steel by more than 5. Suppose that both toughness distributions are normal.

a. Assuming that \({\sigma _1} = 1.2\) and \({\sigma _2} = 1.1\), test the relevant hypotheses using \(\alpha = .001\).

b. Compute \(\beta \) for the test conducted in part (a) when \({\mu _1} - {\mu _2} = 6.\)

The accompanying summary data on compression strength (lb) for \({\bf{12 \times 10 \times 8}}\)in. boxes appeared in the article "Compression of Single-Wall Corrugated Shipping Containers Using Fixed and Floating Test Platens" (J. Testing and Evaluation, 1992: 318-320). The authors stated that "the difference between the compression strength using fixed and floating platen method was found to be small compared to normal variation in compression strength between identical boxes." Do you agree? Is your analysis predicated on any assumptions?

\(\begin{array}{*{20}{c}}{ Method }&{\begin{array}{*{20}{c}}{ Sample }\\{ Size }\end{array}}&{\begin{array}{*{20}{c}}{ Sample }\\{ Mean }\end{array}}&{\begin{array}{*{20}{c}}{ Sample }\\{ SD }\end{array}}\\{ Fixed }&{10}&{807}&{27}\\{ Floating }&{10}&{757}&{41}\\{}&{}&{}&{}\end{array}\)

The accompanying data consists of prices (\$) for one sample of California cabernet sauvignon wines that received ratings of 93 or higher in the May 2013 issue of Wine Spectator and another sample of California cabernets that received ratings of 89 or lower in the same issue.

\(\begin{array}{*{20}{c}}{ \ge 93:}&{100}&{100}&{60}&{135}&{195}&{195}&{}\\{}&{125}&{135}&{95}&{42}&{75}&{72}&{}\\{ \le 89:}&{80}&{75}&{75}&{85}&{75}&{35}&{85}\\{}&{65}&{45}&{100}&{28}&{38}&{50}&{28}\end{array}\)

Assume that these are both random samples of prices from the population of all wines recently reviewed that received ratings of at least 93 and at most 89 , respectively.

a. Investigate the plausibility of assuming that both sampled populations are normal.

b. Construct a comparative boxplot. What does it suggest about the difference in true average prices?

c. Calculate a confidence interval at the\(95\% \)confidence level to estimate the difference between\({\mu _1}\), the mean price in the higher rating population, and\({\mu _2}\), the mean price in the lower rating population. Is the interval consistent with the statement "Price rarely equates to quality" made by a columnist in the cited issue of the magazine?

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