The Genetics 8 IVF Institute developed its XSORT method to...

Name: Problem 19, Chapter 6: Elementary Statistics
Uploaded: 2023-04-02T17:02:25-08:00
Duration: 3 min 51 s
Channel: Prabhakar Kumar
Description: The Genetics 8 IVF Institute developed its XSORT method to increase the probability of conceiving a girl. Among 574 women using that method, 525 had baby girls. Assuming that the method has no effect so that boys and girls are equally likely. find the probability of getting at least 525 girls among 574 babies. Does the result suggest that the XSORT method is effective? Why or why not?

The Genetics 8 IVF Institute developed its XSORT method to increase the probability of conceiving a girl. Among 574 women using that method, 525 had baby girls. Assuming that the method has no effect so that boys and girls are equally likely. find the probability of getting at least 525 girls among 574 babies. Does the result suggest that the XSORT method is effective? Why or why not?

Key Concepts

Binomial Distribution

The binomial distribution is a probability model used to describe the number of successes in a fixed number of independent trials, where each trial has the same probability of success. In this context, each birth is considered a trial with a success (having a girl) probability of 0.5 under the null hypothesis, allowing us to calculate the expected distribution of outcomes if the method has no effect.

Normal Approximation

When dealing with a large number of trials, the binomial distribution can be approximated by the normal distribution using the Central Limit Theorem. This approximation simplifies calculations for tail probabilities, such as determining the probability of observing an extreme outcome like having at least 525 girls out of 574 births under the assumption of no effect.

Hypothesis Testing

Hypothesis testing is a statistical framework used to decide whether there is enough evidence to reject a null hypothesis. In this problem, the null hypothesis is that the XSORT method has no effect on the probability of conceiving a girl, meaning the chance is 50%. The observed data is then compared to the expected distribution to assess the effectiveness of the method.

p-value

The p-value represents the probability of obtaining an outcome as extreme, or more extreme, than the observed result, assuming the null hypothesis is true. A very small p-value indicates that the observed outcome is highly unlikely under the assumption of no effect, which may lead to rejecting the null hypothesis. In this case, calculating the p-value helps determine if the unusually high number of girls suggests that the XSORT method is effective.

Transcript

00:01 In this question, we are going to discuss about the method given in the problem, which is basically for the increment of probability of conceiving a girl.

00:09 So here we can say that population mean will be n times p, where p is basically the probability of having a girl.

00:16 So that is 1 divided with 2 because it is given that probability of having a girl and boy is equally likely, then what will be the value of probability? so it is here going to be q, that is the probability of having a boy, so it will be 1 minus p and that is also equals to 1 divided with 2.

00:33 So n value, that is sample size in our problem is 7, sorry, it is 5, 74 and hence it will be here 575 multiplied with 1 divided with 2 and that is here equals to 287.

00:48 So solving this very expression further, we can say that population standard deviation can be written as, n p q and that is here equals to under root of n value is 574 p value is here 1 divided with 2 and then q value is here also 1 divided with 2 so it will be raised to the power 2 so this value is here going to come around 11 .979 now using the z score formula the z value can be calculated as x minus mu divided with sigma so it will be here 524 point it is here 5 minus it will be 287 divided with it will be here 11 .979.

01:33 So this value is here going to come around 19 .83.

01:39 So from here we can say that the value of probability such that let us here write the probability of x greater than equals to 525 and that is here equals to probability of x and that will be greater than 524 .5 and this is here equals to probability of z and that will be greater than 19 .83 and hence we can say that it will be here equals to 1 minus probability of z and that will be here less than 19 .83 and this is here equals to 1 minus 0 .99 and then here it is 99.

02:20 So this will be here equals to 0 .00 and then here it is 1.

02:26 So from this very value we can say that since the value of probability that we have find out, right? so this was the required probability.

02:34 So here we can say that required probability, p -r -o -b -a -b -i -l -i -t -y -e -e -e -e equals to 0 .00 and then it will be here 0 and now it is here 1...

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