Confidence Intervals for One Mean: Sigma Not Known (t Method)

[@shahsalman3846]

you are simply the best sir ... you've got an art of teaching ...great videos ... brief and concise ..... things that i haven't understood in 3 hrs of stats lecture at the university ... i've just grasped it in 9:45 mins .......just wow

[@jbstatistics]

You are welcome! I'm very glad you find my videos helpful!

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[@jbstatistics]

Thank you for the kind words!

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You're welcome. This is the best place to come for stats help :)

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my report is tomorrow and i wasn't able understand the difference between the z-distribution and the t-distribution. but now, after seeing this video i've understand everything

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[@jbstatistics]

You are welcome. Best of luck on your final exam!

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[@jbstatistics]

You are very welcome Tia! I'm very glad I could be of help.

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[@jbstatistics]

You are very welcome, and thanks very much for the compliment!

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[@jbstatistics]

Or get ahead even! :)

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[@jbstatistics]

You're very welcome Kristen!

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[@jbstatistics]

I'm glad to be of help!

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[@jbstatistics]

You are very welcome!

[@sivanit3150]

Very helpful.... Please sir keep posting more videos on this topic ✨🙌

[@ScilexGuitar]

Cant help but read SE(X(bar)) as Sex bar lol

[@AndresRamirez-fi5uw]

i was just about to comment that

[@ScilexGuitar]

xD

[@simonpaisley101]

Are there any videos on how to find the value at 7:20 using R showing the formula to get 2.447? Thanks in advance!

[@jbstatistics]

That value is the 97.5th percentile of the t distribution with 6 degrees of freedom. The R command qt(.975,6) yields that value.

[@simonpaisley101]

Nevermind! I think I found it. Is it using qt(0.025, 6)?

[@simonpaisley101]

@@jbstatistics Thanks again!!

[@natashasthilaire1950]

5:30 does anyone know how I would find the T-value for this using a TI-84?

[@sahilshah325]

Awesome explanation. Thank you very much

[@jbstatistics]

You are very welcome! Thanks for the compliment!

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you're the stat king 😍👑 thanks ! 😘

[@jbstatistics]

You are very welcome!

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Thanks a lot you just helped me in my mid term review

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YOU ARE THE BEST

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your videos really helping me to pass the Random signal theory course :) Really Thank you =)

[@jbstatistics]

Thanks!

[@SkateB96]

can anyone find the 2.447 in the t tables, i dont know what to look up in the t tables to get it

[@stryfus]

you have to look up the value of (n-1) on the t-table with respect to the value of alpha, so on the team table you should go to 6 since (7-1), and 0.025, which gives you 2.447

[@benandnathan15]

@@stryfus thank you so much I was confused on this and now I get it I was using the confidence interval with the df to find t but now I see what I was doing wrong

[@stryfus]

@@benandnathan15 my pleasure dude, glad I helped you out!

[@SuperNaicker]

Excellent lecture. One question. How do we calculate the same parameters for log normal distribution?

[@StephenLongofono]

Long shot here, but I have a request: Any chance you might do a video on moment generating functions and their applications?

[@jbstatistics]

There's every chance of that, just not in the immediate future :) In the coming year, I'll see what sort of dent I can make in the long list of videos I'd like to make.

[@Maha_s1999]

Waiting eagerly :-D in the mean (pun intended) time.... Merry Xmas from Australia Prof Balka!

[@Andrew6James]

Isn't the sample standard deviation over n-1 to make it an unbiased estimator for the population standard deviation?

[@themaj1x]

Thanks, really helped me out alot!

[@chandnimelwani5630]

Hey Prof, thank you so much for your videos! They are indeed immensely helpful. I was hoping to ask this one question: for alpha = 0.05 and DF = 6, t = 2.447 BUT when alpha = 0.025, DF = 6, t = 2.969. Therefore would t value be 2.969 on each side and thus t(alpha/2) which is t(0.025) = 2.969?

[@PTamu]

Yes the t value for a 0.025 and 6 degrees of freedom would be 2.9692.

[@ismaelrabhironaldo510]

Thanks a million for this video

[@jbstatistics]

You are very welcome.

[@estherlsy]

you're amazing.

[@jbstatistics]

Thanks!

[@iris.y8762]

you r incredible!!

[@mmaking8664]

How do we know that 95% confidence level will lie at 2 sigma while 90% will lie at 1 sigma?

[@ScilexGuitar]

What if we made several set of samples and calculated the confidence interval for each one of them. Would we find a narrower interval with the same confidence level of 95 % in that case?

[@jbstatistics]

I'm not sure what you're asking. If we obtain several samples, and obtain a confidence interval for each of them, them each one would have its own width. The sample size, variance, and confidence level determine the width of the interval.

[@dadir2010]

I ScilexGuitar

[@ScilexGuitar]

Most often the sample standard deviation is not given. How do I calculate it from sample data?

[@jbstatistics]

I discuss measures of variability (including the sample standard deviation) in this video: kzbin.info/www/bejne/eamVpXqLr8ppZtU

[@ScilexGuitar]

Thank you! Much clearer. Do you have any videos of determining confidence intervals for differences between lets say, two populations A and B in some aspect. For example if u test each population for who has a drivers licence. You ask some of each population and thus know the estimated mean of A and B for who can drive, and u wanna construct a confidence interval for the difference between them?

[@jameshandolias8752]

what if your not given mu? and you need to find mu in order to caculate t?

[@dianemonn688]

How did you +- 16.46? Your video has help soo much!

[@maryannking3248]

Can you make a video when a sample deviation is not known and to help find x, or explain what x might be or how to get x?

[@jbstatistics]

I'm not sure what you're asking for here. By the time we get to the topic of this video, what X represents has long since been established. I have many videos that go into the details of the earlier material. Cheers.

[@maryannking3248]

I was looking for a video about sample standard deviation and I did find it. Thank you for the response.

[@soughtafter6455]

Thanks for this, God bless

[@jbstatistics]

You are very welcome!

[@stefanop1340]

Hey man, Thanks very much for the videos! They have really helped me! I would like to ask you a question. In the video you have taken the sample sigma (s) and put it in the formula. But what if we don't have the sample sigma? Is it possible to resolve this problem without it? If i have understood well the sample sigma is equal to sigma (of the population) divided by the square root of "n". So how can I calculate the sample sigma in this case? Some data are missed, am I right? I would really appreciate your help and Thanks again

[@jbstatistics]

Hi Stefano. There's no such thing as a "sample sigma". You may be referring to the sample standard deviation, which is in fact used in this video. sigma divided by the square root of n is the true standard deviation of the sampling distribution of x bar. If you haven't gotten this figured out yet, you may want to check out my video on the sampling distribution of the sample mean at kzbin.info/www/bejne/p2aTeKOKmauMr6s. Cheers.

[@manuelsojan9093]

if your degrees of freedom is not listed on a table, do you round down or round to the nearest integer. For ex., if my degrees of freedom is like 39, do i round down to 30 or to the closest integer, 40?

[@jbstatistics]

I tell my students that I don't much care about that, but your professor might. As a general rule of thumb, we err on the side of being conservative (rounding down), rather than saying we have more DF than we do. (But one could certainly make an argument for choosing the closest value.)

[@manuelsojan9093]

+jbstatistics ok thanks!

[@Denjeanification]

Hi! I love your videos, they have helped me so much but I have a question. In my notes it says that the interval is: sample mean +/- t x s/(square root of n - 1). So is it n - 1 or just n? Thanks in advance! :)

[@joudy5767]

honestly idk

[@cp24ballnews24]

its just n. Am I 7 years to late?

[@A-R53]

What about if we don’t have the standard division either ? How can I find the S or standard division?

[@jbstatistics]

If it's not given to you, then you'll need to calculate it from the data. Calculating the standard deviation is a completely different discussion, and I go into it in my "measures of variability" video.

[@A-R53]

jbstatistics thanks for your answer! I am supposed to calculate the confidence Intervals for a certain data while both the sigma and standard division are unknown!

[@jbstatistics]

@@A-R53 Yes, that's pretty standard. You'll just have to calculate them yourself. You can use software to do it (e.g. R, Excel), or your calculator likely has pre-programmed functions that will do it for you. If all else fails, you can do the grunt work and calculate them from the formulas. Best of luck!

[@A-R53]

jbstatistics thank you for the all effort you put into your videos🌹🌸🌸 They are very helpful :)

[@kharma1111]

Nice video

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[@jbstatistics]

Thanks!

[@fluffyboiix3]

i need help asap

[@imstephruiz]

How the hell did you get 2.447??

[@jbstatistics]

"...and for that, we're going to need to go to software or a t table. And if we go to software or a t table, we can find that that value is 2.447."

[@imstephruiz]

Where can i find the t table, and do i take that number and multiply it to the 0.05 then divide it by 2?

[@chase.2595]

stats god

[@akshatgarg6635]

better than khan academy