Exam 14: Learning From Experiment Data

Correct Answer:

Verified

a) The hypotheses would have to be worded in terms of treatment means or
proportions rather than population means or proportions.
b) For experiments, the individuals or objects must be randomly assigned to
treatments, whereas in testing hypotheses using data from sampling, the samples
must be randomly selected.
For experiments, the number of individuals in the treatment groups must be large
or the distributions of the response variable would need to be approximately
normal in very large treatment groups. For data from sampling, the assumption is
that the samples are large or the population distributions are approximately
normal.
c) Conclusions would need to be worded in terms of treatment means or proportions.
If random sampling from a population preceded the random assignment to
treatments, it would be reasonable to generalize the conclusions about treatment
effects to the population.

Correct Answer:

Verified

Either a one-tailed or two-tailed test is appropriate.
For the one-tailed test, the alternative hypothesis is that the mean for the arthroscopic
physical function scores is greater than the mean for the non-arthroscopic physical
function scores.
a) Let $\mu _ { a }$ represent the mean physical function score for the arthroscopic treatment
$\mu _ { n }$ represent the mean physical function score for the non-arthroscopic treatment
$$\begin{array} { l } H _ { 0 } : \mu _ { a } - \mu _ { n } = 0 \\H _ { a } : \mu _ { a } - \mu _ { n } > 0\end{array}$$ b)
The associated P-value is 0.0568.
c) Since the P-value is greater than .05, we do not have sufficient evidence to
conclude that the arthroscopic surgery results in a greater mean physical function
score than physical therapy and medicine.
For the two-tailed test, the alternative hypothesis is that the mean arthroscopic
physical function score is different from the mean non-arthroscopic physical function
scores.
a) Let $\mu _ { a }$ represent the mean for the arthroscopic treatment
$\mu _ { n }$ represent the mean for the non-arthroscopic treatment
$$\begin{array} { l } H _ { 0 } : \mu _ { a } - \mu _ { n } = 0 \\H _ { a } : \mu _ { a } - \mu _ { n } \neq 0\end{array}$$ b)
The associated P-value is 0.1137.
c) Since the P-value is greater than .05, there is no statistically significant difference
between the treatment means. We do not have sufficient evidence to conclude
that the mean physical function score for arthroscopic surgery treatment is
different from the mean physical function score of the physical therapy and
medicine treatment.

Correct Answer:

Verified

Let $\mu _ { d }$ represent the mean difference $( \mathrm { A } - \mathrm { B } )$ in time to wings level for treatments $\mathrm { A }$ and $\mathrm { B }$
$$\begin{array} { l } H _ { 0 } : \mu _ { d } = 0 \\H _ { a } : \mu _ { d } \neq 0\end{array}$$
The problem states that the assumptions for the $t$ -procedure are met.
$$\begin{array} { l } t = \frac { \bar { d } - 0 } { \sqrt { \frac { s _ { d } ^ { 2 } } { n _ { d } } } } = \frac { - 0.2625 - 0 } { \sqrt { \frac { 2.9747 ^ { 2 } } { 8 } } } = - 0.2496 \\P - \text { value } = 0.81 , d f = 7\end{array}$$
Since the $P$ -value is very large, we cannot reject the null hypothesis. There is not sufficient evidence of a difference in mean recovery times to wings level between drinks $\mathrm { A }$ and $\mathrm { B }$ .