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. An article reported on an experiment involving 90 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 2.5.
a. Does it appear that true average pain level for the control condition exceeds that for the treatment condition by more than 1 at a significance level of 0.01? Carry out a test of appropriate hypotheses using significance level of .01
b. Does it appear that the true average pain level for the control condition exceeds that for the treatmnet condition by more than 1?Carry out a test of appropriate hypotheses

Answer 1

Answer:

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

$\sigma=2.5$

And the statistic is given by this formula:

$z=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where z follows a normal standard distribution

The system of hypothesis on this case are:

Null hypothesis: $\mu_c \leq \mu_t+1$

Alternative hypothesis: $\mu_c >\mu_t+1$

Or equivalently:

Null hypothesis: $\mu_c - \mu_t \leq 1$

Alternative hypothesis: $\mu_c-\mu_t>1$

Our notation on this case :

$n_c =45$ represent the sample size for group control

$n_t =45$ represent the sample size for group  treatment

$\bar X_c =5.2$ represent the sample mean for the group control

$\bar X_t =3.1$ represent the sample mean for the group treatment

And now we can calculate the statistic:

$z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087$  

And now we can calculate the p value using the alternative hypothesis:

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.