It is, quite frankly, embarrassing that you can explain this better than my professor who requires an hour to explain the same subject and yet still fails to get across the point. They get paid to do their job and they can't even do it properly. Many thanks for the assistance.
FOr those of you confused, we are using the t distribution BECAUSE the sample n is less than 30, the standard deviation is 80 and is actually in the video...In other words, it is known
[t_(n-1)] uses the Degrees of Freedom, whereas the [t_a/2] uses the probability. Other than using different values, there probably isn't any difference between the two as you need both values to find the t-value on the t distribution table. The use of either one probably just depends on how you were taught stats (ie. whichever one your teacher uses)
80 is the sample's standard deviation and not the population's. If the population's standard deviation is known, then the formula for the confidence interval is similar to the one shown in the video except that "t alpha/2" is replaced by "z alpha/2"
I'm sorry I'm confused. I do not understand how the title pertains to this video. You advertised that this was to be used when the standard deviation is unknown yet in the example you gave the standard deviation was given (80)?
This video is old but thank you so much for explaining this! I really appreciate it. :) Explanations and examples can be difficult to follow through just text. Have a good day. xx
superb sir...please send link introduction of chi-square test and when it is used....differnces between t-test, f-test and chi-square....waiting for video....
cool video, but your title is misleading. I was expecting an example where the standard deviation is unknown...however in your example the standard deviation is already given; 80.
Hello. The question goes as follows: On the basis of results obtained from a random sample of 100 flights for a company the 95% confidence interval for the mean number of unoccupied seats is given by [12.2; 13.1]. Find sample mean and sample standard deviation. I've come up with the following equation, i know upto here it is right but i don't know how to solve it now. x±1.96 s/ sqr100=12.2, 13.1 Please i'll be grateful if someone can help me ASAP !
Sorry , 10 years late . Sample mean is the mid point of confidence interval = (12.2+13.1)/2 = 12.65 Margin of Error = Z . SD/sqr (n) . Margin of error = difference of Confidence Interval = 0.9 Therefore , 0.9= Z .SD / 10 . Z = 1.96 ( for 95% CI) Solving for SD we get 4.59
Thank you so much! BTW, i think there is a small error in your notes on your website, cos you write that you need the marin of error is determined by several factors such as 2. population standard deviation, but then in your video you say its the sample s.d.
I spent hours trying to figure this out from an unhelpful textbook and online course material and you made it so easy! Thank you!!!
It is, quite frankly, embarrassing that you can explain this better than my professor who requires an hour to explain the same subject and yet still fails to get across the point. They get paid to do their job and they can't even do it properly. Many thanks for the assistance.
FOr those of you confused, we are using the t distribution BECAUSE the sample n is less than 30, the standard deviation is 80 and is actually in the video...In other words, it is known
i see. thank you fo6the clarification. i was confused the standard deviation in this note with the previous notes.
Thanks man, still helping people 5 years in the future!
8 years and still on
@@MoonLight-fs6gl 9 years
10 years! #Keep_it_going
10 lmao
11 years and going strong 💪
10 years later and you saved my life , I appreciate this vidoe
It made sense...? IT MADE SENSE!
this is the go-to explanation on this topic. seen a bunch of other vids and they werent as clear and concise.
Thank you for your contribution ! It is clear. I will apply it right now to my experimental data to get the CI.
Thanx dude...my teacher isn't unable to teach it so simple!
tmrw i got this statistic exam and now i just found the video?! anyway thanks for a great explainantion mr.statistic. :)
THANKS MAN HOW ABOUT IF THE VARIANCE AND STANDARD DEVIATION IS NOT GIVEN BUT YOU ARE ONLY GIVEN SAMPLE DATA
[t_(n-1)] uses the Degrees of Freedom, whereas the [t_a/2] uses the probability. Other than using different values, there probably isn't any difference between the two as you need both values to find the t-value on the t distribution table. The use of either one probably just depends on how you were taught stats (ie. whichever one your teacher uses)
9 years and Still helping
12 years and still helping
Best video I've seen on the subject
Population standard deviation was 80. Said so right from the beginning. How do you find it if that value is unknown?
80 is the sample's standard deviation and not the population's. If the population's standard deviation is known, then the formula for the confidence interval is similar to the one shown in the video except that "t alpha/2" is replaced by "z alpha/2"
happy birthday Dr.E
Thank you. That was a great explanation.
I'm sorry I'm confused. I do not understand how the title pertains to this video. You advertised that this was to be used when the standard deviation is unknown yet in the example you gave the standard deviation was given (80)?
This is when we don't know population standard deviation but 80 is sample standard deviation and denoted by s
Very misleading
It's supposed to be 2.064 for 24 degrees of freedom and 0.025
population standard deviation, sigma, is unknown
80 refers to the sample standard deviation,s
This video is old but thank you so much for explaining this! I really appreciate it. :) Explanations and examples can be difficult to follow through just text. Have a good day. xx
Thank you so much!
Thanks man this really helps!
I think the value of t would be 1.96 in case. Of 95% confidence interval
Amazing explanation!!!!!! Great Job. Thumbs Up!
thank you so much sir. really helped
Thank you very much this video helped me a lot
The t- value is 2.064
superb sir...please send link introduction of chi-square test and when it is used....differnces between t-test, f-test and chi-square....waiting for video....
Thanks--great video.
so to do confidence interval when the standard deviation is unknown,I'll need a distribution table?
cool video, but your title is misleading. I was expecting an example where the standard deviation is unknown...however in your example the standard deviation is already given; 80.
Doug Salinas
population standard deviation is unknown however 80 is the one of sample standard deviation
It is for sample Standard Deviation
thank you help a lot!
but standard deviation 80 is given though
That's just assuming we have the sample standard deviation rather than the population one
@@allaamrauf8214 this comment saved my life i would like you to know that
This is for same Standard Deviation that is why it is denoted by s
thanks u very much!!!
THANK YOUUU
Hello.
The question goes as follows:
On the basis of results obtained from a random sample of 100 flights for a company the 95% confidence interval for the mean number of unoccupied seats is given by [12.2; 13.1]. Find sample mean and sample standard deviation.
I've come up with the following equation, i know upto here it is right but i don't know how to solve it now.
x±1.96 s/ sqr100=12.2, 13.1
Please i'll be grateful if someone can help me ASAP !
Sorry , 10 years late . Sample mean is the mid point of confidence interval = (12.2+13.1)/2 = 12.65
Margin of Error = Z . SD/sqr (n) . Margin of error = difference of Confidence Interval = 0.9
Therefore , 0.9= Z .SD / 10 .
Z = 1.96 ( for 95% CI)
Solving for SD we get 4.59
Thank you so much! BTW, i think there is a small error in your notes on your website, cos you write that you need the marin of error is determined by several factors such as 2. population standard deviation, but then in your video you say its the sample s.d.
Great video. So if the sample size was higher than 25 would we go to the z score table instead?
insted of [t_a/2] my book use [t_(n-1)]
What is the diference, why is that?
why in my table is 1,711 and not 2,0639 ?
also, my teacher gave us the formula where includes t(1-alfa/2) and not alfa/2 ... :S
sorry are you sure about t value you have got? i mean 2.0639
how did you find t=2.0639?
Using T-Table