I'm glad I could help. I'm not sure how communications work on UA-cam but if you have other areas that you need videos for, let me know. I would sure make more for you. Thanks for the comment!
Thank you so much. I don’t know what kind of example this lesson was giving but this helped me entirely! Thank you. I even learned how to use the normal standard table that was never even taught 🤷🏽♀️ the example from my class was ridiculously over done that made it very confusing. Again thank you, you’re awesome😊
I have a question: why, when one is testing the difference between two proportions, does one take a weighted mean estimate of the sample proportions to find the test value for the difference of the true proportions distribution but when one uses the confidence interval formula for the difference between two proportions, one estimates the margin of error by using just the sample proportions without taking their weighting average?
Hi I don't understand how to find ic 95 in this kind of exercise. could you help me? In the article, the difference in 28-day mortality between the Lopinavir / Ritonavir group and the Standard Care group was -5.8%. Mortality was rated as similar in both groups. Among the following fictitious results, which one or which will lead us to conclude that the mortality between the two groups is similar? (The 95% confidence interval is given in square brackets) Result A: Mortality difference: -16% [ -30%; -2%] Result B: Mortality difference: -5.8% [-11%; -0.6%] Result C: Mortality difference: -2% [-17%; 13%] Result D: Mortality difference: -0.5% [-3%; 2%]
Is there a link or pdf to this z-table? Because for whatever reason, the z-tables I find online aren't as accurate as this one and it keeps skewing my results.
That surprising. I figured all z tables were the same. Maybe I need to redo this video with current resources. I can't publish a pdf of what I used because it would break copywrite. I can tell you it comes from author Bluman and is called "Elementary Statistics A step by step Aproach"
@@danielschaben yeah, all the ones I’m seeing has different values in different places than on yours and it keep causing me to get the wrong answers. Even the one in my textbook. But when I used yours through this video, it worked perfectly and I got the correct answer
@@jadalong9587 check your tables. Make sure you are using the same one that has the same area under the normal curve as mine. Sounds like you are on the wrong table type
i have question sir which i was not able to solve my mind got blast but i didnt find satisfied answer question is A random sample of 700 oranges was taken from a large segment and 100 found to be defective construct the 95℅ confidence limit for ℅ of number of oranges in segment
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes. 1. Construct a 90% confidence interval for the true average amount of time customers spent in the restaurant? 2. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant? 3. Dscuss why the answers in Parts (1) and (2) are different? 4. With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less? Entire solution
It's been a while since I've taught this, but essentially it is the percent error we arbitrarily accept as ok. I will think through a better explanation
this video is more helpful than the ones I found in khan academy. you just saved my grades
You're right...I understand this video better as well... life saver 😭😭❣️😘
So glad it was helpful
Yeah true😂👍👍
Something is clearly wrong with this damn textbook for my class lmaooo. This was 10x better than what my teacher is teaching smh
Glad I can help. If there are things you can't find, let me know. I will make videos if your having problems with things
Thank you! This helped so much 😊
This was the best breakdown of this question I've seen this far on the internet.
Wow. Thanks for the comment. I need to get back into making these videos. Been a while since I've made any
youre probably the only one that explained this so well. THANK YOUUUU SO MUCHHHHHHH
Thanks for the comment!
thank you so much!!! saved my life
THANK YOU THANK YOU THANK YOU THANK YOU! I HAVE BEEN BREAKING MY SKULL TO DO THIS AND THIS VIDEO IS AMAZING!!!!!!!!!!!!!
I'm glad I could help. I'm not sure how communications work on UA-cam but if you have other areas that you need videos for, let me know. I would sure make more for you. Thanks for the comment!
Clear make it easier . thank you!
Thank you so much. I don’t know what kind of example this lesson was giving but this helped me entirely! Thank you. I even learned how to use the normal standard table that was never even taught 🤷🏽♀️ the example from my class was ridiculously over done that made it very confusing. Again thank you, you’re awesome😊
Glad I could help.
Very helpful explanation, thank you so much!
Keep up the good work!!!i gota test tomorw...this simple thing really helpd me...thankx
I have a question: why, when one is testing the difference between two proportions, does one take a weighted mean estimate of the sample proportions to find the test value for the difference of the true proportions distribution but when one uses the confidence interval formula for the difference between two proportions, one estimates the margin of error by using just the sample proportions without taking their weighting average?
Exactly what I needed. Thanks
Your welcome
You're
thank you
Awesome video. Thank you.
Thanks for the comment. Glad it was useful!
thanks mate
Hi I don't understand how to find ic 95 in this kind of exercise. could you help me?
In the article, the difference in 28-day mortality between the Lopinavir / Ritonavir group and the Standard Care group was -5.8%. Mortality was rated as similar in both groups. Among the following fictitious results, which one or which will lead us to conclude that the mortality between the two groups is similar? (The 95% confidence interval is given in square brackets) Result A:
Mortality difference: -16% [ -30%; -2%]
Result B:
Mortality difference: -5.8% [-11%; -0.6%]
Result C: Mortality difference:
-2% [-17%; 13%]
Result D: Mortality difference: -0.5% [-3%; 2%]
so clear thank you
You're welcome
Thank you!
You're welcome
Thanks! :)
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Why is it when i download the Z tables i cant find .0450? Have they changed the Z tables since now or ??
yeah but how do you find the equation?
Is there a link or pdf to this z-table? Because for whatever reason, the z-tables I find online aren't as accurate as this one and it keeps skewing my results.
That surprising. I figured all z tables were the same. Maybe I need to redo this video with current resources. I can't publish a pdf of what I used because it would break copywrite. I can tell you it comes from author Bluman and is called "Elementary Statistics A step by step Aproach"
@@danielschaben yeah, all the ones I’m seeing has different values in different places than on yours and it keep causing me to get the wrong answers. Even the one in my textbook. But when I used yours through this video, it worked perfectly and I got the correct answer
@@jadalong9587 check your tables. Make sure you are using the same one that has the same area under the normal curve as mine. Sounds like you are on the wrong table type
Daniel the thumps are distracting
You have nailed it?
I don't know. Trying for sure
Sir quick question how did u get the answer 0.024??
I plugged that zscore times the standard deviation into a calculator. Plug that mess in and see if you get the same thing. Thanks for the comment
why did you take alpha=90 ? I thought 90 = 1 - alpha, shouldnt alpha be 10 ?
he's bored. his z table values r wrong tooo. it has to be 10 yea
i have question sir which i was not able to solve my mind got blast but i didnt find satisfied answer question is
A random sample of 700 oranges was taken from a large segment and 100 found to be defective construct the 95℅ confidence limit for ℅ of number of oranges in segment
Here goes. Sorry been sick. the z-score for 95% confidence is 1.96. The standard error is sqrt((1/7)(6/7)/700) = 0.0132....
(1/7) - 1.96(0.0132)
Thank you very much sir but which formula to so I can understand it and resolve it without reference to learn
So we are 95% confident that the true proportion of bad oranges is between 0.12 and 0.17.
The formula i used is at 4:41 of this video and around 7:00. 4:41 is the general formula. 7:00 is the formula applied to the example I gave.
Thank you very much sir
Do you mind if i keep asking you for solution of problem that which I can't understand
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes. 1. Construct a 90% confidence interval for the true average amount of time customers spent in the restaurant? 2. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant? 3. Dscuss why the answers in Parts (1) and (2) are different? 4. With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less? Entire solution
Why did you use z sub .45? It is a/2, and since confidence was .9, alpha'd be .1. Hence, wouldn't you use .05?
Matt Maloney I was thinking the same thing.
Thanks for the explanation Dan! Great video!
Matt Maloney ?
?
Matt Maloney did you understand why? I have the same question in mind
No clue
please some one help me how to solve below question I am not getting it from so many days
What is alpha???
It's been a while since I've taught this, but essentially it is the percent error we arbitrarily accept as ok. I will think through a better explanation
Not .045, it's .450
could you get me a time stamp? I cant find where your meaning