/*! 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} Q29 E The article "Effect of Internal ... [FREE SOLUTION] | 91影视

91影视

The article "Effect of Internal Gas Pressure on the Compression Strength of Beverage Cans and Plastic Bottles" (J. of Testing and Evaluation, \(1993: 129 - 131\)) includes the accompanying data on compression strength (lb) for a sample of\(12 - oz\)aluminum cans filled with strawberry drink and another sample filled with cola. Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a\(P\)-value. What assumptions are necessary for your analysis?

\(\begin{array}{l}BeverageSampleSizeSampleMeanSampleSD\\Strawberrydrink1554021\\Cola1555415\end{array}\)

Short Answer

Expert verified

Assumption of normality;

Reject null hypothesis.

Step by step solution

01

Find the normal distribution

Denote with\({\mu _1}\)the true average compression strength (lb) for srawberry drink and with\({\mu _2}\)for cola drink. The hypotheses of interest are\({H_0}:{\mu _1} - \)\({\mu _2} = 0\)versus\({H_a}:{\mu _1} - {\mu _2} < 0\). The answer should be based on the\(P\)value.

Given two normal distributions, the random variable (standardized)

\(T = \frac{{\bar X - \bar Y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{S_1^2}}{m} + \frac{{S_2^2}}{n}} }}\)

has approximately students\(t\)distribution with degrees of freedom\(\nu \), where\(\nu \)is

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

has to be rounded down to the nearest integer.

The two-sample\(t\)test for testing\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)uses the following value of test statistic

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

For the adequate alternative hypothesis the adequate area under the \({t_\nu }\) curve is calculated is the \(P\) value.

02

Find the lower tailed test

In order to use this statistic the assumption of normally had to be made. There is enough information to calculate the \(t\) value as

\(\begin{array}{l}t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{540 - 554}}{{\sqrt {\frac{{{{21}^2}}}{{15}}} + \frac{{{{15}^2}}}{{15}}}} = \frac{{ - 14}}{{\sqrt {29.4 + 15} }}\\ = - 2.1\end{array}\)

and the corresponding degrees of freedom of the statistic can be computed using formula given above

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{44.4}^2}}}{{\frac{{{{29.4}^2}}}{{14}} + \frac{{{{15}^2}}}{{14}}}} = \frac{{1971.36}}{{77.81}} = 25.36\)

and round down to the nearest integer to get the degrees of freedom

\(\nu = 25\)

The \(P\) value for the lower tailed test is the probability

\(P = P(T < 2.1) = 0.023\)

where the value can be found in the appendix of the book and statistic \(T\) is students statistic with \(25\) degrees of freedom. Since

\(P = 0.023 < 0.05\)

reject the null hypothesis

From the data you can conclude that extra corbonation of cola results in a higher average compression strength with significance level \(\alpha = 0.05\$ .\)

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

Damage to grapes from bird predation is a serious problem for grape growers. The article "'Experimental Method to Investigate and Monitor Bird Behavior and Damage to Vineyards" (Amer. \(J\). of Enology and Viticulture, 2004: 288-291) reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time (sec) spent on a single visit to the location.

\(\begin{array}{l}Species Location n \overline x SEmean\\Blackbirds Exptl 65 13.4 2.05 \\Blackbirds Natural 50 9.7 1.76\\Silvereyes Exptl 34 49.4 4.78\\Silvereyes Natural 46 38.4 5.06\end{array}\)

a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location.

b. Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses.

c. Estimate the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural

Refer back to the previous exercise.

a. By far the most frequently tested null hypothesis when data is paired is\({H_0}:{\mu _D} = 0\). Is that a sensible hypothesis in this context? Explain.

b. Carry out a test of hypotheses to decide whether there is compelling evidence for concluding that on average diagnosis occurs more than\(\;25\)months after the onset of symptoms.

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores (鈥淰egetarianism, Bone Loss, and Vitamin D: A Longitudinal Study in Asian Vegans and Non-Vegans,鈥 European J. of Clinical Nutr., 2012: 75鈥82)

Diet sample sample sample

Size mean SD

\(\overline {\underline {\begin{array}{*{20}{l}}{ Vegan }&{88}&{5.10}&{1.07}\\{ Omnivore }&{93}&{5.55}&{1.10}\\{}&{}&{}&{}\end{array}} } \)

Calculate and interpret a \(99\% \) \(CI\) for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores (the cited article included a \(95\% \)\(CI\)). (Note: The article described a more sophisticated statistical analysis for investigating bone density loss taking into account other characteristics (鈥渃ovariates鈥) such as age, body weight, and various nutritional factors; the resulting CI included 0, suggesting no diet effect.

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?

a. Show for the upper-tailed test with \({\sigma _1}\) and \({\sigma _2}\)known that as either\(m\) or\(n\) increases, \(\beta \)decreases when \({\mu _1} - {\mu _2} > {\Delta _0}\).

b. For the case of equal sample sizes \(\left( {m = n} \right)\)and fixed \(\alpha \),what happens to the necessary sample size \(n\) as \(\beta \) is decreased, where \(\beta \) is the desired type II error probability at a fixed alternative?

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.