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Chapter 9: Q3 E (page 371)

Pilates is a popular set of exercises for the treatment of individuals with lower back pain. The method has six basic principles: centering, concentration, control, precision, flow, and breathing. The article 鈥淓fficacy of the Addition of Modified Pilates Exercises to a Minimal Intervention in Patients with Chronic Low Back Pain: A Randomized Controlled Trial鈥� (Physical Therapy, \(2013:309 - 321\)) reported on an experiment involving \(86\) subjects with nonspecific low back pain. The participants were randomly divided into two groups of equal size. The first group received just educational materials, whereas the second group participated in \(6\) weeks of Pilates exercises. The sample mean level of pain (on a scale from \(0\) to \(10\)) for the control group at a \(6\)-week follow-up was \(5.2\) and the sample mean for the treatment group was \(3.1\); both sample standard deviations were \(2.3\).

a. Does it appear that true average pain level for the control condition exceeds that for the treatment condition? Carry out a test of hypotheses using a significance level of \(.01\) (the cited article reported statistical significance at this a, and a sample mean difference of \(2.1\) also suggests practical significance)

b. Does it appear that true average pain level for the control condition exceeds that for the treatment condition by more than \(1\)? Carry out a test of appropriate hypotheses

Short Answer

Expert verified

a) Yes, There is enough data to suggest that the genuine average pain level in the control condition is higher than in the treatment condition.

b) Yes, At the \(0.01\) significance level, there is insufficient evidence to support the assertion that the genuine average pain level for the control condition surpasses that for the treatment condition by more than one.

Step by step solution

01

Given information

a)

\(\begin{array}{l}{\rm{Sample means}}, {{\bar x}_1} = 5.2, {{\bar x}_2} = 3.1\\{\rm{Sample standard deviation}}: {s_1} = {s_2} = 2.3\\{\rm{Sample size}}: {n_1} = {n_2} = \frac{{86}}{2} = 43\end{array}\)

\(\alpha = 0.01\)

02

Writing hypothesis

We can employ the \(z\)-test because the samples are huge ( \(n > 30\)). (instead of a t-test).

Assumption made: Exceeds.

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} > {\mu _2}\end{array}\)

03

Finding test statistic

Determine the value of the test statistic:

\(\begin{array}{c}z = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{5.2 - 3.1}}{{\sqrt {\frac{{2.{3^2}}}{{43}} + \frac{{2.{3^2}}}{{43}}} }}\\ \approx 4.23\end{array}\)

04

Finding P-value

If the null hypothesis is true, the P-value is the probability of getting a result more extreme or equal to the standardized test statistic \(z\). Using the normal probability table, determine the probability.

\(\begin{array}{c}P = P(Z > 4.23)\\ = 1 - P(Z < 4.23)\\ \approx 1 - 1\\ = 0\end{array}\)

The null hypothesis \(\alpha \)is rejected if the P-value is less than the alpha significance level.

\(P < 0.01 \Rightarrow Reject {H_0}\)

There is enough data to suggest that the genuine average pain level for the control condition is higher than for the treatment condition.

05

Writing hypothesis

Assumption made: Exceeds \(1\)

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = 1\\{H_a}:{\mu _1} - {\mu _2} > 1\end{array}\)

06

Finding test statistic

Determine the value of the test statistic:

\(\begin{array}{c}z = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{(5.2 - 3.1) - 1}}{{\sqrt {\frac{{2.{3^2}}}{{43}} + \frac{{2.{3^2}}}{{43}}} }}\\ \approx 2.22\end{array}\)

07

Finding P-value

If the null hypothesis is true, the P-value is the probability of getting a result more extreme or equal to the standardized test statistic \(z\). Using the normal probability table, determine the probability.

\(\begin{array}{c}P = P(Z > 2.22)\\ = 1 - P(Z < 2.22)\\ = 1 - 0.9868\\ = 0.0132\end{array}\)

The null hypothesis \(\alpha \)is rejected if the P-value is less than the alpha significance level.

\(P > 0.01 \Rightarrow Fail to reject {H_0}\)

There is enough data to suggest that the genuine average pain level for the control condition is higher than for the treatment condition.

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To decide whether two different types of steel have the same true average fracture toughness values, n specimens of each type are tested, yielding the following results:

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Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article 鈥淒omestic Fuels, Indoor Air Pollution, and Children鈥檚 Health鈥� (Annals of the N.Y. Academy of Sciences, \(2008:209 - 217\)) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (\(LPG\)). For the \(755\) children in biomass households, the sample mean peak expiratory flow (a person鈥檚 maximum speed of expiration) was \(3.30L/s\), and the sample standard deviation was \(1.20\). For the \(750\) children whose households used liquefied petroleum gas, the sample mean \(PEF\) was \(4.25\) and the sample standard deviation was \(1.75\).

a. Calculate a confidence interval at the \(95\% \) confidence level for the population mean \(PEF\) for children in biomass households and then do likewise for children in \(LPG\) households. What is the simultaneous confidence level for the two intervals?

b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average \(PEF\) is lower for children in biomass households than it is for children in \(LPG\) households (the cited article included a P-value for this test)

c. \(FE{V_1}\), the forced expiratory volume in \(1\) second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEV1 was \(2.3L/s\) and the sample standard deviation was \(.5L/s\). If this information is used to compute a \(95\% \) \(CI\) for population mean \(FE{V_1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain

In medical investigations, the ratio \(\theta = {p_1}/{p_2}\)is often of more interest than the difference \({p_1} - {p_2}\)(e.g., individuals given treatment 1 are how many times as likely to recover as those given treatment\(2?)\). Let\(\hat \theta = {\hat p_1}/{\hat p_2}\). When \(m\)and n are both large, the statistic \(ln(\theta )\)has approximately a normal distribution with approximate mean value \(ln(\theta )\)and approximate standard deviation \({[(m - x)/(mx) + (n - y)/(ny)]^{1/2}}.\)

  1. Use these facts to obtain a large-sample 95 % CI formula for estimating\(ln(\theta )\), and then a CI for \(\theta \)itself.
  2. Return to the heart-attack data of Example 1.3, and calculate an interval of plausible values for \(\theta \)at the 95 % confidence level. What does this interval suggest about the efficacy of the aspirin treatment?

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

Consider the pooled\(t\)variable

\(T = \frac{{(\bar X - \bar Y) - \left( {{\mu _1} - {\mu _2}} \right)}}{{{S_p}\sqrt {\frac{1}{m} + \frac{1}{n}} }}\)

which has a\(t\)distribution with\(m + n - 2\)df when both population distributions are normal with\({\sigma _1} = {\sigma _2}\)(see the Pooled\(t\)Procedures subsection for a description of\({S_p}\)).

a. Use this\(t\)variable to obtain a pooled\(t\)confidence interval formula for\({\mu _1} - {\mu _2}\).

b. A sample of ultrasonic humidifiers of one particular brand was selected for which the observations on maximum output of moisture (oz) in a controlled chamber were\(14.0, 14.3, 12.2\), and 15.1. A sample of the second brand gave output values\(12.1, 13.6\),\(11.9\), and\(11.2\)("Multiple Comparisons of Means Using Simultaneous Confidence Intervals," J. of Quality Technology, \(1989: 232 - 241\)). Use the pooled\(t\)formula from part (a) to estimate the difference between true average outputs for the two brands with a\(95\% \)confidence interval.

c. Estimate the difference between the two\(\mu \)'s using the two-sample\(t\)interval discussed in this section, and compare it to the interval of part (b).
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