Question

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 “Efficacy of the Addition of Modified Pilates Exercises to a Minimal Intervention in Patients with Chronic Low Back Pain: A Randomized Controlled Trial” (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

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

Given information

a)

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

\(\alpha = 0.01\)

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.

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

Finding test statistic

Determine the value of the test statistic:

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

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.

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.

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

Finding test statistic

Determine the value of the test statistic:

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

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.