91Ó°ÊÓ

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

$H_0:P_1=P_2$

$H_{\mathrm{Î\pm }}:P_1>P_2$

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

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.

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

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.

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

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.

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

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.