Question

Acupuncture and pregnancy A study reported in the medical journal Fertility and Sterility sought to determine whether the ancient Chinese art of acupuncture could help infertile women become pregnant. One hundred sixty healthy women who planned to have IVF were recruited for the study. Half of the subjects $\left(80\right)$were randomly assigned to receive acupuncture $25$minutes before embryo transfer and again $25$minutes after the transfer. The remaining $80$women were assigned to a control group and instructed to lie still for $25$minutes after the embryo transfer. Results are shown in the table below.

	Acupuncture	Control group
Pregnant	34	21
Not pregnant	46	59
Total	80	80

Is the pregnancy rate significantly higher for women who received acupuncture? To find out, researchers perform a test of $H_0:P_1=P_2$versus $H_{\alpha };P_1>P_2$, where$P_1\&P_2$are the actual pregnancy rates for women like those in the study who do and don't receive acupuncture, respectively.

(a) Name the appropriate test and check that the conditions for carrying out this test are met.

(b) The appropriate test from part (a) yields a P-value of $0.0152$Interpret thisP-value in context.

(c) What conclusion should researchers draw at the $\alpha =0.05$significance level? Explain.

(d) What flaw in the design of the experiment prevents us from drawing a cause-and-effect conclusion? Explain.

Short answer

(a) Two sample $z$test can be use.

(b) The p value of $0.0152$indicates that there is a $1.52$percent chance that there will be no difference in pregnancy rates between the treatment and control groups.

(c) Pregnancies are more common among women who receive acupuncture than among women who do not.

(d) The women in the control group were aware that they had not had acupuncture since they could see it being done to them. This may have an impact on their behaviour, influencing whether or not they become pregnant. This is when the experiment fails.

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that

The number of women in the treatment group who became pregnant was$=34$

In the control group, the number of women who became pregnant was lower.$=21$

total number of women in treatment group$=80$

In the control group, there were a total of $80$women.

$P_1$s the percentage of women in the treatment group who became pregnant.be the proportion of women who became pregnant in the control group $P_2$

The null and alternative hypothesis are

$H_0:P_1=P_2$

$H_{\alpha }:P_1>P_2$

We have to name the appropriate test and check that the conditions for carrying out this test are met.

Part (a) Step 2: Explanation

To execute a two-sample z test for proportional differences, the following conditions must be met.

1. The sample should be random

2.$np>10$

3.$n\left(1-p\right)>10$

4.The observation should be independent.

$$\begin{gathered} P_1=\frac{34}{80} \\ =0.43 \end{gathered}$$

$$\begin{gathered} P_2=\frac{21}{80} \\ =0.26 \end{gathered}$$

$$\begin{gathered} n_1{p}_1=80*0.43 \\ =34 \\ {n}_1{p}_1>10 \\ {n}_1\left(1-p_1\right)=80*(1-0.34) \\ =46 \\ {n}_1\left(1-p_1\right)>10 \end{gathered}$$

$$\begin{gathered} n_2{p}_2=80*0.26 \\ =21 \\ {n}_2{p}_2>10 \\ {n}_2\left(1-p_2\right)=80*(1-0.26) \\ =59 \\ {n}_2\left(1-p_2\right)>10 \end{gathered}$$

The women's groups are treated as independent because the experiment is randomized. If one woman becomes pregnant, no information about another woman is revealed. As a result, each group's individual observations are likewise deemed independent.

Therefore two sample$z$test can be use.

Part (a) Step 1: Given Information 

Given in the question that the p value is$0.0152$we have to interpret p value of the significance test.

Part (b) Step 2: Explanation  

The p value is the probability that the null hypothesis is correct. The p value of $0.0152$ indicates that there is a $1.52$percent chance that there will be no difference in pregnancy rates between the treatment and control groups.

Part (c)  Step 1: Given Information  

Given in the question that the value of p is $0.0152$and the significance level is$\alpha =0.05$. we have to know the conclusion drawn n significance test at $5\%$ level of significance.

Part (c) Step 2: Explanation  

The p value of $0.0152$is less than the $0.05$level of significance. The null hypothesis is rejected at the level of significance of $5$percent. The proportion of pregnancies in the treatment group is higher than in the control group, we conclude. As a result, the proportion of pregnancies among women who receive acupuncture is higher than among women who do not.

Part (d)  Step 1: Given Information  

We have to find out that at what flaw in the design of the experiment prevents us from drawing a cause-and-effect conclusion.

Part (d) Step 2: Explanation  

The women in the control group were aware that they had not had acupuncture since they could see it being done to them. This may have an impact on their behaviour, influencing whether or not they become pregnant. This is when the experiment fails.