Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution.

\(A\)

\({\bf{B}}\)

\(26 - < 30\)

6

4

\(30 - < 34\)

12

9

\(34 - < 38\)

15

19

\(38 - < 42\)

7

10

\(m = 40\)

\(m = 42\)

Compute a 95 % CI for the difference between the true proportions of all specimens of alloys A and B that have an ultimate strength of at least\(34ksi\).

Short Answer

Expert verified

The 95% confidence interval is from -0.35 to 0.07.

Step by step solution

01

Step 1: To compute a 95 % CI for the difference between the true proportions of all specimens of alloys a and B

A confidence interval for \({p_1} - {p_2}\)with confidence level of approximately \(100(1 - \alpha )\% \)is

\({\hat p_1} - {\hat p_2} \pm {z_{\alpha /2}} \times \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \)

The confidence interval can be used when\(m \times {\hat p_1},m \times {\hat q_1},n \times {\hat p_1}\), and\(n \times {\hat q_1}\)are at least 10.

The proportion of specimens of alloys A that have an ultimate strength of at least \(34{\rm{ksi}}\) is

\({\hat p_1} = \frac{{15 + 7}}{{40}} = 0.55\)

The proportion of specimens of alloys B that have an ultimate strength of at least \(34{\rm{ksi}}\) is

\({\hat p_1} = \frac{{19 + 10}}{{40}} = 0.69.\)

02

Finding confidence interval

For the 95 % confidence interval \(\alpha = 0.05\)and\({z_{\alpha /2}} = {z_{0.025}} = 1.96\). Thus, the \(95\% \)confidence interval is

\(\begin{array}{c}{{\hat p}_1} - {{\hat p}_2} \pm {z_{\alpha /2}} \times \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \\ = 0.55 - 0.69 \pm 1.96 \times \sqrt {\frac{{0.55 \times 0.45}}{{40}} + \frac{{0.69 \times 0.31}}{{42}}} \\ = - 0.14 \pm 1.96 \times 0.106\\ = ( - 0.35,0.07).\end{array}\)

The 95% confidence interval for the difference between the two proportions is from -0.35 to 0.07

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Researchers sent 5000 resumes in response to job ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2500 of them had "white sounding" first names, such as Brett and Emily, whereas the other 2500 had "black sounding" names such as Tamika and Rasheed. The resumes of the first type elicited 250 responses and the resumes of the second type only 167 responses (these numbers are very consistent with information that appeared in a Jan. 15. 2003, report by the Associated Press). Does this data strongly suggest that a resume with a "black" name is less likely to result in a response than is a resume with a "white" name?

The invasive diatom species Didymosphenia geminata has the potential to inflict substantial ecological and economic damage in rivers. The article "Substrate Characteristics Affect Colonization by the BloomForming Diatom Didymosphenia geminata" (Acquatic Ecology, 2010:\(33 - 40\)) described an investigation of colonization behavior. One aspect of particular interest was whether the roughness of stones impacted the degree of colonization. The authors of the cited article kindly provided the accompanying data on roughness ratio (dimensionless) for specimens of sandstone and shale.

\(\begin{array}{*{20}{l}}{ Sandstone: }&{5.74}&{2.07}&{3.29}&{0.75}&{1.23}\\{}&{2.95}&{1.58}&{1.83}&{1.61}&{1.12}\\{}&{2.91}&{3.22}&{2.84}&{1.97}&{2.48}\\{}&{3.45}&{2.17}&{0.77}&{1.44}&{3.79}\end{array}\)

\(\begin{array}{*{20}{l}}{ Shale: }&{.56}&{.84}&{.40}&{.55}&{.36}&{.72}\\{}&{.29}&{.47}&{.66}&{.48}&{.28}&{}\\{}&{.72}&{.31}&{.35}&{.32}&{.37}&{.43}\\{}&{.60}&{.54}&{.43}&{.51}&{}&{}\end{array}\)

Normal probability plots of both samples show a reasonably linear pattern. Estimate the difference between true average roughness for sandstone and that for shale in a way that provides information about reliability and precision, and interpret your estimate. Does it appear that true average roughness differs for the two types of rocks (a formal test of this was reported in the article)? (Note: The investigators concluded that more diatoms colonized the rougher surface than the smoother surface.)

The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of \(43\) chronic schizophrenics, resulting in \(\bar x = 2.69\) and \({s_1} = 2.30,\), as well as for \(45\) normal subjects, resulting in \(\bar y = 6.35\) and \({s_2} = 4.03.\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha = .01\)

. (Hint: \({H_0}\) and \({H_a}\) here have a different form from the three standard cases. Let \({\mu _1}\) and \({\mu _2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta = 2{\mu _1} - {\mu _2}\). Write \({H_0}\) and \({H_a}\) in terms of \(\theta \), estimate \(\theta \), and derive \({\hat \sigma _{\tilde \theta }}\) (โ€œReduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients,โ€ Nature, July 28, 1972: 225โ€“226).) \(\alpha = .01\)

McNemar's test, developed in Exercise 56, can also be used when individuals are paired (matched) to yield n pairs and then one member of each pair is given treatment 1 and the other is given treatment 2 . Then \({X_1}\)is the number of pairs in which both treatments were successful, and similarly for\({H_0}\)\({X_2},{X_3}\), and\({X_4}\). The test statistic for testing equal efficacy of the two treatments is given by\(\left( {{X_2} - {X_3}} \right)/\sqrt {\left( {{X_2} + {X_3}} \right)} \), which has approximately a standard normal distribution when \({H_0}\)is true. Use this to test whether the drug ergotamine is effective in the treatment of migraine headaches.

The data is fictitious, but the conclusion agrees with that in the article "Controlled Clinical Trial of Ergotamine Tartrate" (British Med, J., 1970: 325-327).

Reconsider the data of Example \(9.6\), and calculate a \(95\% \)upper confidence bound for the ratio of the standard deviation of the triacetate porosity distribution to that of the cotton porosity distribution.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free