Chapter 10, Problem 10.67RE

(a)

To determine

To Explain: that this was an observational study or an experiment and justifies the answer.

Answer to Problem 10.67RE

Experiment

Explanation of Solution

DEFINATIONS

An experiment is imposing some treatment on individuals in order to observe their responses.

An observational study is trying to collect information without any disturbance of the scan they are observing.

The study is an experimental, because the given assigned treatments are the acupuncture treatments and the control treatments.

(b)

To determine

To Explain: that there is convincing evidence of a relationship between pregnancy status and whether acupuncture was administered and carry out a significance test and report the conclusion.

Answer to Problem 10.67RE

There is enough evidence to help the claim that there is a connection between pregnancy status and whether acupuncture was administered.

Explanation of Solution

Formula used:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ E=np\end{array}$

Calculation:

Assume:

  $\alpha =0.05$

The null hypothesis statement is that the variables are not connected, where the alternative hypothesis statement is that they are connected.

  $H_0:$ There is no connection between treatment and pregnancy.

  $H_1:$ There is a connection between treatment and pregnancy.

The row and column totals are

	Acupuncture	Control	Total
Pregnant	34	21	55
Not pregnant	46	59	105
Total	80	80	160

The expected frequencies $E$ are

  $\begin{array}{c}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{80\times 55}{160}=27.5\\ E_{12}=\frac{r_1\times c_2}{n}=\frac{80\times 105}{160}=52.5\\ E_{21}=\frac{r_2\times c_1}{n}=\frac{80\times 55}{160}=27.5\\ E_{22}=\frac{r_2\times c_2}{n}=\frac{80\times 105}{160}=52.5\end{array}$

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(34-27.5\right)}^2}{27.5}+\frac{{\left(21-52.5\right)}^2}{52.5}\\ +\frac{{\left(46-27.5\right)}^2}{27.5}+\frac{{\left(59-52.5\right)}^2}{52.5}\\ =33.6866\end{array}$

The degree of freedom is

  $df=\left(r-1\right)\left(c-1\right)=\left(2-1\right)\left(2-1\right)=1$

The P-Value is the probability of getting the value of the test statistic, or a value more extreme and row $df=1$ .

  $P<0.001$

If the P-value is equal or less than to the significance level, then the null hypothesis is rejected:

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is enough evidence to help the claim that there is an connection between pregnancy status and whether acupuncture was administered.

(c)

To determine

To Explain: the conclusion that acupuncture caused an increase in pregnancy.

Answer to Problem 10.67RE

These subjects are aware of which treatment they get and the results may then be affected by the placebo effect

Explanation of Solution

These subjects are aware of which treatment they get and the results may then be affected by the placebo effect. The placebo effect is an effect build by a placebo, which cannot be due to the placebo itself, and is therefore reason by the patient’s belief in that treatment. Since the subjects are of which treatment they get, the control group would not have a placebo effect, where the acupuncture group would have a placebo effect.