/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q26 E The article "The Influence of Co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q26 E (page 380)

The article "The Influence of Corrosion Inhibitor and Surface Abrasion on the Failure of Aluminum-Wired Twist-On Connections" (IEEE Trans. on Components, Hybrids, and Manuf. Tech., 1984: 20-25) reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying SAS output suggest that the true average potential drop for alloy connections (type 1) is higher than that wfor EC connections (as stated in the article)? Carry out the appropriate test using a significance level of .01. In reaching your conclusion, what type of error might you have committed?

(Note: SAS reports the\(P\)-value for a two-tailed test.)

\(\begin{array}{*{20}{c}}{ Type }&{}&N&{ Mean }&{ Std Dev }&{ Std Error }\\1&{20}&{17.49900000}&{0.55012821}&{0.12301241}&{}\\2&{20}&{16.90000000}&{0.48998389}&{0.10956373}&{}\\{}&{ Variances }&T&{ DF }&{Prob > |T|}&{}\\{ Unequal }&{3.6362}&{37.5}&{0.0008}&{}&{}\\{}&{ Equal }&{3.6362}&{38.0}&{0.0008}&{}\end{array}\)

Short Answer

Expert verified

Reject the null hypothesis.

Step by step solution

01

Testing the hypotheses

Let\({\mu _1}\)denotes the true average potential drop for alloy connections and\({\mu _2}\)denotes the true average for the EC connections.

The hypotheses of interest are\({H_0}:{\mu _1} - {\mu _2} = 0\)versus\({H_a}:{\mu _1} - {\mu _2} > 0\).

Testing this hypothesis, the results provided in the exercise, when the variances are unequal (see exercise output), is

\(t = 3.6362,df = 37.5\).

The pvalue for two tailed test is 0.0008.

The the p- value for the upper tailed test can be computed as

\(P = \frac{1}{2} \cdot 0.0008 = 0.0004\)

Since

\(\begin{array}{c}P = 0.0004\\ < 0.01\end{array}\)

reject the null hypothesis at 1% level of significance.

The type I error could have been made- rejecting the null hypothesis when it is true.

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 91影视!

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

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether \((S)\)or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows:

Before

After

S

F

S

\({{\bf{X}}_{\bf{1}}}\)

\({{\bf{X}}_{\bf{2}}}\)

F

\({{\bf{X}}_{\bf{3}}}\)

\({{\bf{X}}_{\bf{4}}}\)

Where\({{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + }}{{\bf{x}}_{\bf{3}}}{\bf{ + }}{{\bf{x}}_{\bf{4}}}{\bf{ = n}}\). Let\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}\), and \({p_4}\)denote the four cell probabilities, so that \({p_1} = P(S\) before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased.

a. State the two hypotheses of interest in terms of\({p_1},{p_2}\),\({p_3}\), and \({p_4}\).

b. Construct an estimator for the after/before difference in success probabilities

c. When n is large, it can be shown that the rv \(\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}{{\bf{X}}_{\bf{j}}}} \right){\bf{/n}}\) has approximately a normal distribution with variance given by\(\left[ {{{\bf{p}}_{\bf{i}}}{\bf{ + }}{{\bf{p}}_{\bf{j}}}{\bf{ - }}{{\left( {{{\bf{p}}_{\bf{i}}}{\bf{ - }}{{\bf{p}}_{\bf{j}}}} \right)}^{\bf{2}}}} \right]{\bf{/n}}\). Use this to construct a test statistic with approximately a standard normal distribution when \({H_0}\)is true (the result is called McNemar's test).

d. If\({{\bf{x}}_{\bf{1}}}{\bf{ = 350,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = 150,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = 200}}\), and\({x_4} = 300\), what do you conclude?

Recent incidents of food contamination have caused great concern among consumers. The article "How Safe Is That Chicken?" (Consumer Reports, Jan. 2010: 19-23) reported that 35 of 80 randomly selected Perdue brand broilers tested positively for either campylobacter or salmonella (or' both), the leading bacterial causes of food-borne disease, whereas 66 of 80 Tyson brand broilers tested positive.

  1. Does it appear that the true proportion of noncontaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level .01.
  2. If the true proportions of non-contaminated chickens for the Perdue and Tyson brands are .50 and .25, respectively, how likely is it that the null hypothesis of equal proportions will be rejected when a .01 significance level is used and the sample sizes are both 80?

According to the article "Modeling and Predicting the Effects of Submerged Arc Weldment Process Parameters on Weldment Characteristics and Shape Profiles" (J. of Engr. Manuf., \(2012: 1230 - 1240\)), the submerged arc welding (SAW) process is commonly used for joining thick plates and pipes. The heat affected zone (HAZ), a band created within the base metal during welding, was of particular interest to the investigators. Here are observations on depth\((mm)\)of the HAZ both when the current setting was high and when it was lower.

\(\begin{array}{*{20}{l}}{ Non - high }&{1.04}&{1.15}&{1.23}&{1.69}&{1.92}\\{}&{1.98}&{2.36}&{2.49}&{2.72}&{}\\{}&{1.37}&{1.43}&{1.57}&{1.71}&{1.94}\\{ High }&{2.06}&{2.55}&{2.64}&{2.82}&{}\\{}&{1.55}&{2.02}&{2.02}&{2.05}&{2.35}\\{}&{2.57}&{2.93}&{2.94}&{2.97}&{}\end{array}\)

a. Construct a comparative boxplot and comment on interesting features.

b. Is it reasonable to use the two-sample \(t\) test to test hypotheses about the difference between true average HAZ depths for the two conditions?

c. Does it appear that true average HAZ depth is larger for the higher current condition than for the lower condition? Carry out a test of appropriate hypotheses using a significance level of . \(01\).

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.

Anorexia Nervosa (AN) is a psychiatric condition leading to substantial weight loss among women who are fearful of becoming fat. The article "Adipose Tissue Distribution After Weight Restoration and Weight Maintenance in Women with Anorexia Nervosa" (Amer. J. of ClinicalNutr., 2009: 1132-1137) used whole-body magnetic resonance imagery to determine various tissue characteristics for both an AN sample of individuals who had undergone acute weight restoration and maintained their weight for a year and a comparable (at the outset of the study) control sample. Here is summary data on intermuscular adipose tissue (IAT; kg).

Assume that both samples were selected from normal distributions.

a. Calculate an estimate for true average IAT under the described AN protocol, and do so in a way that conveys information about the reliability and precision of the estimation.

b. Calculate an estimate for the difference between true average AN IAT and true average control IAT, and do so in a way that conveys information about the reliability and precision of the estimation. What does your estimate suggest about true average AN IAT relative to true average control IAT?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation
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.