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The accompanying data on response time appeared in the article "'The Extinguishment of Fires Using Low-Flow Water Hose Streams-Part II" (Fire Technology, 1991: 291-320).

Good visibility

.43 1.17 .37 .47 .68 .58 .50 2.75

Poor visibility

1.47 .80 1.58 1.53 4.33 4.23 3.25 3.22

The authors analyzed the data with the pooled t test. Does the use of this test appear justified? (Hint: Check for normality.

The z percentiles for n=8 are -1.53, -.89, -.49, -.15 , .15, .49, .89, and 1.53.)

Short Answer

Expert verified

Any test which requires normality should not be used.

Step by step solution

01

To analyzing the   t test appear justified

It appears that the good visibility does not come from a normally distributed population-based on the normal probability plot given below. For the pooled t test assumption of normality for both populations is needed. However, only the poor visibility satisfies that. Thus,

any test which requires normality should not be used

02

Step 2:To analyzing the   t test appear justified

03

To analyzing the  t test appear justified

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Most popular questions from this chapter

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 (โ€œVegetarianism, Bone Loss, and Vitamin D: A Longitudinal Study in Asian Vegans and Non-Vegans,โ€ European J. of Clinical Nutr., 2012: 75โ€“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 (โ€œcovariatesโ€) such as age, body weight, and various nutritional factors; the resulting CI included 0, suggesting no diet effect.

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.

The article "Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population" (Ergonomics, 1997: 642-649) reported that (-473.13,1691.9) was a large-sample 95 % confidence interval for the difference between true average thenar muscle volume \(\left( {m{m^3}} \right.)\) for sufferers of carpal tunnel syndrome and true average volume for nonsufferers. Calculate and interpret a 90 % confidence interval for this difference.

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?

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

Arsenic is a known carcinogen and poison. The standard laboratory procedures for measuring arsenic \(concentration (\mu g/L)\) in water are expensive. Consider the accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method (from the article "Evaluation of a New Field Measurement Method for Arsenic in Drinking Water Samples," J. of Emvir. Engr., 2008: 382-388).

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 3 19.70 1.10 0.65

2 3 10.90 0.60 0.35

Estimate for difference: 8.800

95% CI for difference: 6.498,11.102

T -Test of difference =0 (vs not = ):

T -Value =12.1 P -Value =0.001 DF=3

What conclusion do you draw about the two methods, and why? Interpret the given confidence interval. (Note: One of the article's authors indicated in private communication that they were unsure why the two methods disagreed.)

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