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Consider the accompanying data on breaking load (\(kg/25\;mm\)width) for various fabrics in both an unabraded condition and an abraded condition (\(^6\)The Effect of Wet Abrasive Wear on the Tensile Properties of Cotton and Polyester-Cotton Fabrics," J. Testing and Evaluation, \(\;1993:84 - 93\)). Use the paired\(t\)test, as did the authors of the cited article, to test\({H_0}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} > 0\)at significance level . \(01\)

Short Answer

Expert verified

Do not reject null hypothesis.

Step by step solution

01

Test statistic value for testing the hypotheses

The Paired \(t\) Test:

When \(D = X - Y\) (difference between observations within a pair), \({\mu _D} = \)\({\mu _1} - {\mu _2}\) and null hypothesis

\({H_0}:{\mu _D} = {\Delta _0},\)

the test statistic value for testing the hypotheses is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }},\)

where\(\bar d\) and \({s_D}\) are the sample mean and the sample standard deviation of differences\({d_i}\), respectively. In order to use this test, assume that the differences\({D_i}\) are from a normal population. Depending on alternative hypothesis, the\(P\) value can be determined as the corresponding area under the \({t_{n - 1}}\) curve.

The hypotheses of interest are\({H_0}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} > 0\). The following table represents the data for unabraded condition and abraded condition, as well as the difference between corresponding observed pairs: \(\begin{array}{*{20}{c}}i&{{\rm{ Unabraded, }}{x_i}}&{{\rm{ Abraded, }}{y_i}}&{{\rm{ Difference, }}{d_i} = {x_i} - {y_i}}\\1&{36.4}&{28.5}&{7.9}\\2&{55}&{20}&{35}\\3&{51.5}&{46}&{5.5}\\4&{38.7}&{34.5}&{4.2}\\5&{43.2}&{36.5}&{6.7}\\6&{48.8}&{52.5}&{ - 3.7}\\7&{25.6}&{26.5}&{ - 0.9}\\8&{49.8}&{46.5}&{3.3}\\{}&{}&{}&{}\end{array}\)

02

Find the sample variance

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean\(\bar d\)for the differences is

\(\bar d = \frac{1}{8} \cdot (7.9 + 35 + \ldots + 3.3) = 7.25\)

Reminder:\({\mu _D}\)is the true average of the difference of breaking load for fabric in unabraded and abraded condition and\({\Delta _0} = 0\). The only missing value to compute the test statistic value is the sample standard deviation of differences\({d_i}\).

The Sample Variance\({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

Where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{\infty } \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation\(s\)is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(s_D^2{\rm{ }} = \frac{1}{{8 - 1}} \cdot \left( {{{(7.9 - 7.25)}^2} + {{(35 - 7.25)}^2} + \ldots + {{(3.3 - 7.25)}^2}} \right) = 140.726\)

and the sample standard deviation\({s_D} = \sqrt {140.726} = 11.8628\)

03

Find the value of P

The test statistic value is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{7.25 - 0}}{{11.8628/\sqrt 8 }} = 1.73\)

The test is one-sided where the alternative hypothesis is\({H_a}:{\mu _D} > 0\), therefore the\(P\)value is the area under the the\({t_{n - 1}} = {t_{8 - 1}} = {t_7}\)curve to the right of\(t\)value

\(P = P(T > 1.73) = 1 - P(T \le 1.73) = 1 - 0.936 = 0.064\)

Where\(T\)has student distribution with\(7\)degrees of freedom, and the probability as computed using software (you can use table in the appendix of the book). At significance level\(0.01\), because

\(P = 0.064 > 0.01\)

do not reject null hypothesis

at given significance level.

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Most popular questions from this chapter

1. An article in the November \(1983\) Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline \(AA\)batteries and Eveready Energizer Alkaline \(AA\) batteries were given as \(4.1\) hours and \(4.5\) hours, respectively. Suppose these are the population average lifetimes.

a. Let \(\bar X\) be the sample average lifetime of \(100\) Duracell batteries and \(\bar Y\) be the sample average lifetime of \(100\) Eveready batteries. What is the mean value of \(\bar X - \bar Y\) (i.e., where is the distribution of \({\bf{\bar X - \bar Y}}\) centered)? How does your answer depend on the specified sample sizes?

b. Suppose the population standard deviations of lifetime are \(1.8\) hours for Duracell batteries and \(2.0\) hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic \(\bar X - \bar Y\), and what is its standard deviation?

c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of \(\bar X - \bar Y\) (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of \(10\) batteries of each type? Explain

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?

The National Health Statistics Reports dated Oct. \(22,2008\), included the following information on the heights (in.) for non-Hispanic white females:

Sample sample Std. Error

Age Size Mean Mean

\(\begin{array}{*{20}{l}}{20 - 39}&{866}&{64.9}&{.09}\\{60 and older }&{934}&{63.1}&{.11}\\{}&{}&{}&{}\end{array}\)

  1. Calculate and interpret a confidence interval at confidence level approximately \(95\% \) for the difference between population mean height for the younger women and that for the older women.
  2. Let \({\mu _1}\) denote the population mean height for those aged \(20 - 39\) and \({\mu _2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \({H_0}:{\mu _1} - {\mu _2} = 1 and {H_a}:{\mu _1} - {\mu _2} > 1,\) and then carry out a test of these hypotheses at significance level \(.001\)
  3. Based on the \(p\)-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.
  4. What hypotheses would be appropriate if \({\mu _1}\) referred to the older age group, \({\mu _2}\) to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than \(1\)in.?

The article "Pine Needles as Sensors of Atmospheric Pollution" (Environ. Monitoring, 1982: 273-286) reported on the use of neutron-activity analysis to determine pollutant concentration in pine needles. According to the article's authors, "These observations strongly indicated that for those elements which are determined well by the analytical procedures, the distribution of concentration is lognormal. Accordingly, in tests of significance the logarithms of concentrations will be used." The given data refers to bromine concentration in needles taken from a site near an oil-fired steam plant and from a relatively clean site. The summary values are means and standard deviations of the log-transformed observations.

Site Sample Size Mean Log Concentration SD of Log Concentration

Steamplant 8 18.0 4.9

Clean 9 11.0 4.6

Let \(\mu _1^*\) be the true average log concentration at the first site, and define \(\mu _2^*\) analogously for the second site.

a. Use the pooled t test (based on assuming normality and equal standard deviations) to decide at significance level .05 whether the two concentration distribution means are equal.

b. If \(\sigma _1^8\) and \(\sigma _2^*\) (the standard deviations of the two log concentration distributions) are not equal, would \({\mu _1} and {\mu _2}\) (the means of the concentration distributions) be the same if \(\mu _1^* = \mu _2^*\) ? Explain your reasoning.

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis:

\(\begin{array}{*{20}{l}}{ - 24}&{ - 12}&{ - 55}&{ - 15}&{ - 30}&{ - 60}&{ - 14}&{ - 21}\\{ - 48}&{ - 12}&{ - 25}&{ - 53}&{ - 61}&{ - 69}&{ - 80}&{}\end{array}\)

a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?

b. Calculate a lower\(95\% \)confidence bound for the population mean difference, and interpret the resulting bound.

c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a\(95\backslash \% \)upper confidence bound for the corresponding population mean difference?

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