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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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Most popular questions from this chapter

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

The article "Evaluating Variability in Filling Operations" (Food Tech., 1984: 51-55) describes two different filling operations used in a ground-beef packing plant. Both filling operations were set to fill packages with \(1400\;g\)of ground beef. In a random sample of size 30 taken from each filling operation, the resulting means and standard deviations were \(1402.24\;g\) and \(10.97\;g\) for operation 1 and \(1419.63\;g\) and \(9.96\;g\) for operation 2.

a. Using a .05 significance level, is there sufficient evidence to indicate that the true mean weight of the packages differs for the two operations?

b. Does the data from operation 1 suggest that the true mean weight of packages produced by operation 1 is higher than\(1400\;g\)? Use a \(.05\)significance level.

Reconsider the data of Example \(9.6\), and calculate a \(95\% \)upper confidence bound for the ratio of the standard deviation of the triacetate porosity distribution to that of the cotton porosity distribution.

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo-controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to GI symptoms ("Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips," J. of the Amer. Med. Assoc., 1998: 150-152). Among 529 individuals in the TG control group, 17.6 % experienced an adverse GI event, whereas among the 563 individuals in the olestra treatment group, 15.8 % experienced such an event.

a. Carry out a test of hypotheses at the 5 % significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control treatment.

b. If the true percentages for the two treatments were 15 % and 20 %, respectively, what sample sizes

\((m = n\)) would be necessary to detect such a difference with probability 90?

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.?
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