My brain was locked up as Iwas trying to understanding all thee concepts but this video unlocked my brain and I understand better now. Than you sir for such a great explanation 🫡
Sir, Great explanations. I have the following questions. When using a country population with gender as the control variable for stratification as follows: Males Strata: compose of Males with decease, Males without decease ( Control group) Female Strata: Women with decease, and Woman without deceased(Control group). I have a few questions 1) I am wondering if I have to calculate the the sample size of the entire population or should I calculate the sample for each strata individually. 2) The problem that I ran into is that according to your explanation N= [(Z*SIGMA)/ d]^2 precision "d" was kept to 0.05, Z=1.645 for Alpha 0.05 and CI=95%, but I do not have SIGMA of the population. SO how do I find SIGMA... is this for SIGMA population or estrata In a different publication I found Ntotal= [Z*/ d]^2 * [ Phat(1-Phat)], where Phat is a predermine value of sensitivity =90% ( or Specificity). Using the second method I am getting Ntotal = [1.645/0.05]^2 * [0.9*0.1]= 354 ( If this the correct TOTAL sample size for my study? I think it is but I want to be sure. and the next question is can I divide my sample size in even stratas... say 177male and 177 female and could I further divide EACH strata in a test and control group evenly say: Males Strata: compose of Males with decease (n= 89) , Males without decease (n =89 for Control group) Female Strata: Women with decease (n =89), and Woman without deceased (n =89 for Control group). So I am confuse above these two approaches 3) I read in the literature that I should consider using this table www.research-advisors.com/tools/SampleSize.htm In my case for a 95 CI and a total population in a country of 10 million, the table indicate to use 384 samples. But I need to know if this is for the entire experiment or for each strata as well... similar questions to the one above. The numbers I calculated for the size of the sample in (2) = (354) are not too different to the ones on the table in (3)= 384. I am opting to use N=400 to opt for potential participants dropping the study. I will truly appreciate your input. Sincerely, Manny
Hello sir, is that a rule of thumb that when sample size is 30 den v used t otherwise z, As I have heard somewhere n some research paper has used t while sample size is more than 30. So how can I know the status and be confident among z and t only on the basis of sample size? How will it affect my study relaibility and validity if I used t instead of z? Waiting for ur expert opinion, Other opinion is also highly appreciated Thanks
1.96 means its a two tailed test, leaving 2.5% in each side of the normal distribution curve. so here the level of confidence is 97.5% at the Z value of 1.96. If it was one tailed test then at 95% confidence level the Z value would have been 1.645
this is the best well-explained video I have come across on this topic. Thank you so much, sir
u r videos are very clear....i am faculty teaching market research and i refer ur videos....keep posting
this is the best well explained video I have come across for this topic..thank you so much sir!!
Very informative! Thanks for the tutorial!
Very good teaching Sir🙏
thanks for sharing the video.
Sir very nice session. Very informative. Thank you so much
My brain was locked up as Iwas trying to understanding all thee concepts but this video unlocked my brain and I understand better now. Than you sir for such a great explanation 🫡
Thank u sir, I'm passing my exam beacause of u this time. Lots of respect.
Very informative ❤️, perfect ok 👍😂
Yayaa 😂😂😂
Pinallla😂🌝
Thank you for your brief explanation!!
This is the reason why IIT is craze in our country. Quality of teachers is incredible
Thank you Dr. Nayak for your lecture useful to me
Sir how we can understand the 2 tail test which condition?
Sir your videos are superb and helped us a lot. We are waiting for your upcoming videos.
Samajh bhi aaya h
Ha bhai
Please explain significant value??
Really nice explain..
Sir, Great explanations.
I have the following questions. When using a country population with gender as the control variable for stratification as follows:
Males Strata: compose of Males with decease, Males without decease ( Control group)
Female Strata: Women with decease, and Woman without deceased(Control group).
I have a few questions
1) I am wondering if I have to calculate the the sample size of the entire population or should I calculate the sample for each strata individually.
2)
The problem that I ran into is that according to your explanation N= [(Z*SIGMA)/ d]^2
precision "d" was kept to 0.05, Z=1.645 for Alpha 0.05 and CI=95%, but I do not have SIGMA of the population. SO how do I find SIGMA... is this for SIGMA population or estrata
In a different publication I found Ntotal= [Z*/ d]^2 * [ Phat(1-Phat)], where Phat is a predermine value of sensitivity =90% ( or Specificity). Using the second method I am getting
Ntotal = [1.645/0.05]^2 * [0.9*0.1]= 354 ( If this the correct TOTAL sample size for my study? I think it is but I want to be sure. and the next question is can I divide my sample size in even stratas... say 177male and 177 female and could I further divide EACH strata in a test and control group evenly say:
Males Strata: compose of Males with decease (n= 89) , Males without decease (n =89 for Control group)
Female Strata: Women with decease (n =89), and Woman without deceased (n =89 for Control group).
So I am confuse above these two approaches
3) I read in the literature that I should consider using this table www.research-advisors.com/tools/SampleSize.htm
In my case for a 95 CI and a total population in a country of 10 million, the table indicate to use 384 samples.
But I need to know if this is for the entire experiment or for each strata as well... similar questions to the one above.
The numbers I calculated for the size of the sample in (2) = (354) are not too different to the ones on the table in (3)= 384. I am opting to use N=400 to opt for potential participants dropping the study.
I will truly appreciate your input.
Sincerely,
Manny
superb explanation...
What is the null and alternative hypothesis in mean and if we select the small or larger one actually i have assignment on this so plz help me
Very informative & useful video.....
Why is he going for 2 tail test when we know that given sample mean for Ha is greater than Ho?
Thank you was useful
Thankyou so much. Very helpful
What is direction of test
Very informative
Subtitles are not needed and are a big hinderance. By the way good video
RUclips has an option to disable it. It is an extra, not compulsory. If you don't like it, disable it.
Very nice explanation sir
Sir u r the best
Nice sir..👍
Can you please elaborate the concept of normal distribution
Some more long distance if video is perfect
Very well explained
Good lecture.
Hello sir, is that a rule of thumb that when sample size is 30 den v used t otherwise z,
As I have heard somewhere n some research paper has used t while sample size is more than 30.
So how can I know the status and be confident among z and t only on the basis of sample size?
How will it affect my study relaibility and validity if I used t instead of z?
Waiting for ur expert opinion,
Other opinion is also highly appreciated
Thanks
They both are same. Just the difference is in theri sample size. There is no Z test in the software as he has already mentioned in video.
Thanks 4 sharing.
Awesome
Why the given example of gpa is two tailed?
Because outcome can be either more or less than 3
Best explanation
Thank u sir
Thank u so much sr ❤️
Please let me know sir if small sample for test is 30 per cent of total population or 30 number of respondents
if population or sample size is less than 30 then t test is used else z test is to be used.
Well explained sir
Thanks to sharing this video
Very clear explanation
Well explained
Nice lecture Sir. Really informative
From where does this z= 1.96 came out, can anyone explain ??
Saurabh Dhiman that is Z critical value ! Which can be found out frm the z table .at 95 percent level of confidence
1.96 means its a two tailed test, leaving 2.5% in each side of the normal distribution curve. so here the level of confidence is 97.5% at the Z value of 1.96. If it was one tailed test then at 95% confidence level the Z value would have been 1.645
Good Effort
❤❤❤
Thank u sir🙋
Just came here to watch how the toppers r studying 😅😢😂😂
good
Good one thank you
i like your vodeos
Thanks boss
nice
Sir help me content acha hai but marketing ki jrurt hai 😭😭😭🙏🙏please marketing kr do mere channel ki.....
There are many mistakes
what are they
What are the mistakes pls tell them
Too gud sir
thanks sir
Thank you sir
Very well explained