Prayer and pregnancy Two hundred women who were about to...

Prayer and pregnancy Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Women in the treatment group were intentionally prayed for by several people (called intercessors) who did not know them, a process known as intercessory prayer. The praying continued for three weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. ${ }^{17}$ Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of $H_{0}: p_{1}=p_{2}$ versus $H_{a}: p_{1}>p_{2},$ where $p_{1}$ and $p_{2}$ are the actual pregnancy rates for women like those in the study who do and don't receive intercessory prayer, 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 0.0007 . Interpret this $P$ -value in context. (c) What conclusion should researchers draw at the $\alpha=$ 0.05 significance level? Explain. (d) The women in the study did not know whether they were being prayed for. Explain why this is important.

Prayer and pregnancy Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Women in the treatment group were intentionally prayed for by several people (called intercessors) who did not know them, a process known as intercessory prayer. The praying continued for three weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. ${ }^{17}$ Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of $H_{0}: p_{1}=p_{2}$ versus $H_{a}: p_{1}>p_{2},$ where $p_{1}$ and $p_{2}$ are the actual pregnancy rates for women like those in the study who do and don't receive intercessory prayer, 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 0.0007 . Interpret this $P$ -value in context.
(c) What conclusion should researchers draw at the $\alpha=$ 0.05 significance level? Explain.
(d) The women in the study did not know whether they were being prayed for. Explain why this is important.

Key Concepts

Blinding in Experimental Design

Blinding refers to keeping study participants unaware of their group assignment, which helps prevent biases that could influence the study’s outcomes. Ensuring that subjects do not know whether they are receiving the treatment or not reduces the risk of placebo effects or differential behavior that could skew the results.

Experimental Design and Random Assignment

Random assignment of subjects to treatment or control groups is crucial because it helps ensure that the groups are statistically similar in all respects except for the treatment. This minimizes bias and confounding factors, making any observed differences between the groups more likely attributable to the treatment effect.

Significance Level and Decision Making

This concept involves using a predetermined threshold (commonly ? = 0.05) to determine the statistical significance of the results. If the p-value is less than the significance level, the null hypothesis is rejected, which suggests that the observed effect (e.g., a higher pregnancy rate in the treatment group) is statistically significant.

P-value Interpretation

The p-value is the probability of obtaining a result at least as extreme as the one observed, assuming that the null hypothesis is true. A small p-value indicates that such an extreme result is unlikely under the null hypothesis, thereby providing evidence to reject it in favor of the alternative hypothesis.

Hypothesis Testing for Two Proportions

This statistical method is used to compare the rates (proportions) of an outcome between two independent groups. It typically involves establishing a null hypothesis that there is no difference between the groups (p1 = p2) and an alternative hypothesis that there is a difference or a directional effect. The test generally employs a z-test under the conditions that both samples are random and sufficiently large so that the normal approximation applies.

Transcript

00:01 In this video, we have been given information about prayer and pregnancy and based on the information we can here write that the null hypothesis will be that p1 is equals to p2 and the alternate hypothesis will be here equals to p1 and then it is greater than p2.

00:20 So we note that we want to execute a test about comparing two population mean that is done by using the two sample z test.

00:30 So there are some conditions that has to be followed before we can use this very, we can say two sample z test.

00:38 So let us check out those conditions that is whether they are satisfied or not.

00:41 So those conditions are namely first one is random condition and it can be written as random.

00:48 So random condition means that the sample data should be randomized.

00:51 There is no selection bias.

00:54 So since the study is a random experiment, so this very condition is satisfied.

01:00 And hence we can write here satisfied.

01:03 Now the second condition is called as 10 % rule.

01:07 So 10 % rule can be written over here that our first sample is of 88 women.

01:15 So these 88 is lesser than 10 % of all the women because in this very study we are considering more than 88 % of 8 % 0 women.

01:25 Now here we have n2 which is equals to basically 81 which is again less than 10 % of all women and so this condition is satisfied.

01:36 Now here we have to check the third condition which is called as normal condition.

01:40 So in this very condition we can say that the value of number of successes and failures is greater than 10 % for each sample because the number of successes can be written as for 10 % percent.

01:54 44 comma 21 and the value of failures can be written as 88 minus 44 that will be 44 and then here it is 81 minus 21 that will be 60 so this very value is greater than 10 % for each sample that's why the condition is satisfied here we can mention that x1 represent number of successes x2 represent number of successes for second sample x1 is for first sample and the number of failure is written as n1 minus x1 in first sample and for second sample it is n2 minus x2.

02:31 So that's how we have written the above value by calculating them.

02:35 So now let us check out the here result.

02:39 So result will be two sample.

02:42 Here it will be like this.

02:43 That is two sample z test can be used and it will be here for the difference in two proportion.

02:52 So it will be for the difference, d -i -f -e -r -e -n -c -e in two proportions.

03:02 So this is the answer for the first part of this very problem.

03:05 Now, we have to find out the answer for other parts as well.

03:09 But first, we can put this thing inside a box in order to highlight it.

03:13 So let us proceed further.

03:16 And here it can be said that in the second part of this very problem, let us write that we are going to have the information that hypothesis test, that is null hypothesis, will be p1 is equal to p2 and then alternate hypothesis, that is ha will be here...

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