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Given in the question that

The number of women in the treatment group who became pregnant was $=44$

In the control group, the number of women who became pregnant was lower. localid="1651964264860" $=21$

$=88$total number of women in treatment group

In the control group, there were a total of $=81$women.

$P_1$is 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.

$$\begin{gathered} P_1=\frac{44}{88} \\ =0.5 \end{gathered}$$

$$\begin{gathered} P_2=\frac{21}{81} \\ =0.26 \end{gathered}$$

$$\begin{gathered} n_1{p}_1=88*05 \\ =44 \end{gathered}$$

$n_1{p}_1>10$

$$\begin{gathered} n_1\left(1-p_1\right)=88*\left(1-0.5\right) \\ =44 \end{gathered}$$

$n_1\left(1-p_1\right)>10$

$$\begin{gathered} n_2{p}_2=81*0.26 \\ =21 \\ {n}_2{p}_2>10 \\ {n}_{2\left(1-p_2\right)}=81*\left(1-0.26\right) \\ =60 \\ {n}_2\left(1-p_2\right)>10 \end{gathered}$$

The experiment is randomized, and the women's groups are treated as independent. 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 used.

Given in the question that the p value is $.0007$we have to interpret the p value of the significance test.

The p value represents the probability that the null hypothesis is correct. The p value of$0.0007$indicates that there is a$0.07$ percent chance that the pregnancy rates of women in the treatment and control groups are identical.

Given in the question that the value of p is $0.007$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 $0.0007$is less than the significance level of $0.05$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 percentage of pregnancies among women who prayed for are higher than those who did not.

We have to know the importance of women in the study not knowing if they were being prayed for.

If the women had realized they were being prayed for, it is possible that their behaviour, whether consciously or unconsciously, might have influenced whether or not they became pregnant.