Yes that is correct as well. As you can tell my point during that section was to show the effect of sample size on standard error and I failed to change the t-critical to 2.01-ish at df=49. Luckily it actually exaggerates the point I was trying to make about standard error. :) I kind of planned that part at the last minute as an extension to the video about how sample size can have a dramatic effect on the conclusion. Thank you again and fix will be forthcoming.
Alpha is also called the "significance level" or Type I error. We can never be 100% confident in an estimation. Alpha is the probability we accept up front of being "wrong" about our estimation. Alpha is used for for any confidence interval, z-distribution or t-distribution. We may get a sample mean say 5% of the time out in the tail(s) of the distribution.
You are very welcome. Confidence intervals appear all throughout introductory statistics. I do not know unfortunately how your professor characterizes a "type" of confidence interval. The ones I routinely work with are single sample with sigma known and unknown. And two-sample with sigmas known and sigmas unknown. I have done videos for the single sample and I am currently working on videos for the two-sample situation.
I have been attempting for days to understand the difference between confidence levels and confidence intervals. Explanations from multiple textbooks did not help me to discern the differences. You cleared the matter up in no time at all. Thank you so much!
Another correction in the example where you have increased the Sample size from 30 to 50 the value of "t" will also change as the degree of freedom has changed from 29 to 49
wew thank you so much, I rewatched this video and the previous one thrice and re-read my class notes a dozen times because i thought my understanding of the concept was totally wrong because I couldn't figure out why the t value was unchanged! I should have just read the comments first :P
Hello! One or two-tailed intervals reflect the direction of the effect we are analyzing. The two-tailed interval has two finite values at its ends. The one-tailed will have either negative infinity or positive infinity at one end depending on the direction. It really just depends on the wording of the problem. Most CIs are two-tailed because having infinity as a boundary does not make much practical sense. The problem or question dictates which is used.
Hi Brandon, thank you for this wonderful course. I was wondering if there is a mistake occuring at the question at around 1 minute, which should be 'the sample standard deviation is $4' instead of 'the sample mean is $4'? Thanks.
12:57 When sample size increased to 50, isn't the shape of the t distribution changed as well? With 49 df rather than 29, the t value should be 2.01, not 2.045 any more
Amazing teaching! I was an A+ student in Stats before, but I literally forgot all the concepts a few years later. This hopefully will help me tremendously towards taking my CFA exam soon!
I think it was a typo. I think he meant the standard deviation of the sample. If you followed the example, you'll find him writing s = 4. s refers to the standard deviation of the sample.
In the confidence interval for the mean, the alpha is divided by 2; a/2. So 95% confidence would have an alpha of 5% or .05. That is divided evenly between both tails. So .025 probability in the lower tail and .025 in the upper tail. If alpha was 10% or .10 then a/2 is .05 probability in the lower tail and .05 probability in the upper tail (90% in the middle).
Brandon, your videos have been true lifesavers for me! Currently enduring BUS308 Statistics for Managers, and the explanations you post ring like lead crystal: I reference each video I use in the assigned discussion forums. Spreading the word!! Thanks again :)
Hi Cherie! Problem #2 was adapted from a textbook I use in my work. So the standard deviation comes from a similar problem. I assume the problem was part of a study of some sort at one time.
Hello! Yes you are correct that is a typo. :( I will place that correction in the description and as an annotation. Also, if you go to my channel page, all my videos are in playlists in their natural order. Each playlist follows the same order a textbook would and inside each playlist the topics are also in order. Thank you so much! - B
Great videos and thank you. I think I found an error in Statistics 101: Confidence interval problems, sigma unknown. In the example of the false customers when the sample size was changed from 30 to 50, the t statistic remained at 2.045; it should have been closer to 2.0.
Regarding Problem #2: I am stumped on how you got an "xbar" of 3661.5. Where did this come from? I have tried so many formulas and can't figure this out. Thank you for your videos. Very helpful!
I understand everything here except how you came up with your t scores when looking at the concern points. For example, when moving the sample to 50 people, how did you derive the t=2.38. i imagine it's a simple answer.
thank you so much. but just a question. in the last example you increased the sample size but you did not change the t statistics (2,045) even if we have a new degree of freedom ??? please answer me if you get my comment
Dr.Brandon , When you changed the Sample Size to 50 instead of 30 , the degrees of freedom changes alspo to 49 and so 2.045 should change to 2.0 instead, am i right?
Brandon, Thank you SO much for doing these videos. They are - hands down - the most clear and easy to understand a all I have come across. I do have a question. In your example in this video on False Customers, where you change the sample size from 30 to 50, why do you only calculate a new estimate for the standard deviation and notalso determine a new t table value based on 49 degrees of freedom? Thank you in advance.
Hi Brandon, One question here. If the standard error of the mean is the standard deviation of the sampling distribution then in the case where we do not know the population standard deviation why don't we calculate the standard error of the mean by calculating the standard deviation of the sampling distribution?? Thanks.
I am having problems with The scores on an examination in finance are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination. 421, 422, 468, 507, 530, 567, 586 Send data to Excel Find a 90 %confidence interval for the population standard deviation. Then complete the table below.
Thank you. My professor mentions 4 types of Confidence Intervals. I am not sure what they are because textbook is not clear. Also need to know the formula for each one - again one runs into the other and I am unable to distinguish them.
Thank you for these videos. I'm reviewing and I'm at a point of confusion. Regarding the example of the show rooming, is this an example of a hypothesis test that will be explained in further videos? Or, is the example not related to hypothesis testing?
You really nailed it, awesome explanation. Now a burning question which I am sure everyone has, how do you obtain and calculate the SAMPLE STANDARD DEVIATION?
@@denissedeo895 Just like the way you calculate Mean of a Sample, calculate the Variance of a Sample and find the Square Root of that Variance. Hope this shed some light to help you finding the Root.
How do you come up with the sample standard deviation 1958 seconds? I have been going insane trying to figure this out! Also, where did the 4320s go? Wouldn't that be n? Someone, anyone HELP please!
1st question : where do 1958 population standard deviation came from? remember, you collect a sample with the size of 30 (30 different people with different bills). from this 30 measurements, you can regularly calculate mean (3661.5 $), variance (sigma(xi-3661.5)^2) divided by 30-1), and standard deviation (square root of variance). you've got to calculate each of 30 samples then substract with the mean, etc. to get the standard deviation. to cut on time, Mr. Brandon merely provides us with the standard deviation first hand since he assumes we all know how to calculate it (see playlist 02 on Mr. Brandon's channel) 2nd question : i don't get exactly what you mean, but 4,320 seconds is just assumption, so the manager of this store makes an assumption, -- *if more than 15% of the worker's average time (4,320 seconds) is spent on false customer* --, then we got a problem. the first samples shows a range of average time spent on customer of 2930-4392 seconds >> so, from this ONE sample, the manager concludes that they got a problem with false customer the second samples (which is larger in size, and therefore more representative) shows a range of average time spent on customer of 3,104 to 4,218 seconds >> notice that the range of average time spent on false customer doesn't reach the 4,320 marks, therefore they've got no problem with the false customer
In the initial slide of example, sample mean is written as $4, instead of sample standard deviation. Next slide is correct. Please rectify/ revise initial one.
Its the concern point (4320) minus the sample mean (3661.5) divided by sample standard deviation (357.48). (4320-3661.5)/357.48 I was very confused too, it took me a minute to figure it out.
These videos have helped so much. I've shared them with others in class. I do have a question about CI that still isn't clear. Why with a binomial distribution do you square root pq/n, when for these problems you only root the n?
@@BrandonFoltz it would be nice to put the x-bar value in the slide as well, so that your viewers would be less confused on small things like this. Overall, your videos are great and super helpful for a struggling student like me. Kudos to you!
Great video! Question: the standard deviation formula is different depending on whether the data is the full population or just a sample. When calculating the standard deviation for a sample to then use in the SEM formula, presumably we should use the sample std dev formula. Is that correct? Or do we say that perhaps since the t-distribution we're going to compare against is already using degrees-of-freedom, perhaps we only do it once?
One caveat to the customers draining resources example. When you increased the sample size from 30 to 50 and held everything constant, that's not to say that the extra 20 observations would have in fact changed the sample mean an thus the sample standard deviation and thus the standard error of the sample and therefore the threshold could have fallen into our confidence interval, right?
you can find the t-value by looking at the t-table or using an excel function. if you're using excel use the t.inv(prob;df) function to find the points. i'll go over the basic a bit here, -why do you need to find t-value? to find the lower and upper point where 95% of the sample distribution falls -what do you need to find t-value? just the fact of how many % of confidence interval (90%, 95%, 97%, etc) do you want to have, and what are your degrees of freedom (size of sample minus 1) -how do you find t-value? on the t-table, just find the degrees of freedom of 29 on the row, and 5% significance on the column, you'll find 2.042, which is exactly the same as the video. on excel, you use the function =t.inv(probability; df) >> =t.inv(probability0.975;29) we use 0.975 (97.5%) because from the left-most (-infinity), to lower point (2.5%), to upper point(97.5%) contains 97.5% probability.
What is the difference between saying that there is 95% probability that the population mean falls whithin an interval and saying that we are 95% confident that this interval contains the population mean?
Hi, I was hoping to find an explanation on how to find the t value for the concern point. so far I know only to use degrees of freedom and the alpha / 2. how do we know the df for the concern point. or if we know the concern points value along the x axis how do we work t point backwards. is this in another video? I don't even know how to search internet for this instance. if you see this question I'd love an answer. Thanks!
Hi Brandon, let’s assume I want to calibrate a lab balance using a reference weight. I measure 30 times the weight and calculate the mean and standard deviation. So I can estimate a confidence interval for the sample mean. After some time I want to check the balance to verify it is still accurate. I take a double (or triple) measurement of the reference weight and obtain an average value. How do I know if the value obtained is within the acceptable range?
Hi Brandon. Thanks heaps for doing this awesome work. I am learning a great deal out of your videos. Just a question.. in the 2nd example when the mean was increased from 30 to 50.. the t distribution through the chart was still 2.045 (which I reckon based on 29) Shouldn't we put 1.96 when the sample size is 50? Thanks
Hi, Great video and explanation. I have small question regarding the example 2 customer service drain, normally when you try to find t value, we use degrees of freedom and alpha/2 = 0.025 but when you found that the concerned point is coming inside the calculated confidence interval,you found the t value as 1.80 at 4320 seconds. How did you get that t value? did you change the alpha value? please assist with your answer. cheers :)
So the difference between the "standard deviation of the sample" and the "standard deviation of the sample means, ie the standard error of the sample" Is that the former tells you the standard deviation of ONE sample, while the later tells you the standard error of the distribution of multiple sample means if you were to take multiple samples?
hello, we find t value based on alpha value at 95 % confidence interval and degrees of freedom. what if i need t value for 93.6 % confidence interval? it is not shown in t distribution table. then how to find it? and what if i know some specific value of sample beyond 95% confidence may be at 96.5 %. how to find t value there? please answer :)
Brandon, I am having major problems with confidence levels. In one example (looking at 95% confidence level) I find that I need alpha divided by 2 (.05) in order to find z score and on another example I find that the confidence level - of 95% has to be divided by 2 to find the z score. I am confused.
Just heard you mention alpha at the beginning of this example. Am I correct to assume that the T distribution table refers to the equivalent of alpha. What does alpha has to do with the t distribution? Maybe that is what I am missing. Thank you.
Hi Brandon, thank you so much for your videos. i have a question though. the standard deviation is 4 (on video 2.23secs) so you put this into an equation and get a standard deviation of 1.03. whats the difference between the sd 4 and the sd 1.03? would be really gratefull if you could come back to me asap. thanks again!
I think there's an error. @1:00 xbar is mentioned instead of s @1:05 s is assumed. I guess it was a typo, mean instead of sample. I love your videos. It would be cool if a playlist was made with the natural order.
in both examples - Why is the sample standard deviation not substituted directly as sigma (X bar)in confidence interval formula ( X bar + 2.145sigma (x bar)) ? The sample standard deviation is divided by square root of Sample size like it was done in population std deviation problem...Why is it not substituted in formula directly...Really confused..Can someone please clarify?
Hello Brandon Foltz This is my first time to go public and ask a simple question? Where do you get X=3661.5 * Is the number random select or a sample number from you? * OR given number that you select? I am just stuck with the X=3661.5 Probability you had shown how to get it but I need to be refresh up or give me some clues or tell me to go to this video and there is where the x=3661.5 Okay Thanks
Quick Question: Is this a mistake the first problem in the wording says that the sample mean is $4, but the next slide it was used as a standard deviation. Is that an error or am I under looking something? Love your videos by the way
Yes that is correct as well. As you can tell my point during that section was to show the effect of sample size on standard error and I failed to change the t-critical to 2.01-ish at df=49. Luckily it actually exaggerates the point I was trying to make about standard error. :) I kind of planned that part at the last minute as an extension to the video about how sample size can have a dramatic effect on the conclusion. Thank you again and fix will be forthcoming.
Hello Sir @Brandon_Foltz
, Can I have the slides of these videos? It will help us to prepare well (jaydebdas111@gmail.com)
Alpha is also called the "significance level" or Type I error. We can never be 100% confident in an estimation. Alpha is the probability we accept up front of being "wrong" about our estimation. Alpha is used for for any confidence interval, z-distribution or t-distribution. We may get a sample mean say 5% of the time out in the tail(s) of the distribution.
You are very welcome. Confidence intervals appear all throughout introductory statistics. I do not know unfortunately how your professor characterizes a "type" of confidence interval. The ones I routinely work with are single sample with sigma known and unknown. And two-sample with sigmas known and sigmas unknown. I have done videos for the single sample and I am currently working on videos for the two-sample situation.
I have been attempting for days to understand the difference between confidence levels and confidence intervals. Explanations from multiple textbooks did not help me to discern the differences. You cleared the matter up in no time at all. Thank you so much!
Another correction in the example where you have increased the Sample size from 30 to 50 the value of "t" will also change as the degree of freedom has changed from 29 to 49
I also noticed this issue.
Good catch. He put a note in the video though. He may have edited it after your comment.
wew thank you so much, I rewatched this video and the previous one thrice and re-read my class notes a dozen times because i thought my understanding of the concept was totally wrong because I couldn't figure out why the t value was unchanged! I should have just read the comments first :P
@@tentathesane8032 same here, relieved i'm not going crazy :)
how does the sample mean remain the same as well when the sample size is increased? is that an assumption?
Hello! One or two-tailed intervals reflect the direction of the effect we are analyzing. The two-tailed interval has two finite values at its ends. The one-tailed will have either negative infinity or positive infinity at one end depending on the direction. It really just depends on the wording of the problem. Most CIs are two-tailed because having infinity as a boundary does not make much practical sense. The problem or question dictates which is used.
Hi Brandon, thank you for this wonderful course. I was wondering if there is a mistake occuring at the question at around 1 minute, which should be 'the sample standard deviation is $4' instead of 'the sample mean is $4'? Thanks.
Perhaps he edited the video, but at 1:18 he repeats that s, the standard deviation of the sample, is 4
Yes， it should be sample standard deviation on the slide.
12:57 When sample size increased to 50, isn't the shape of the t distribution changed as well? With 49 df rather than 29, the t value should be 2.01, not 2.045 any more
That's exactly what popped into my mind also
that's correct.
Amazing teaching! I was an A+ student in Stats before, but I literally forgot all the concepts a few years later. This hopefully will help me tremendously towards taking my CFA exam soon!
How did you get a standard deviation of 4 ? At min 1:18 ......when you said the sample mean is 4
Anyone know? Brandon please answer
I think it was a typo. I think he meant the standard deviation of the sample. If you followed the example, you'll find him writing s = 4. s refers to the standard deviation of the sample.
well fix it BF!
I think it is a printing mistake. It would be sample standard deviation $4 and in question no 2 the value of sample mean $20 is given.
Hi guys, this means the sample is $4 and standard deviation is $4 too correct?
In the confidence interval for the mean, the alpha is divided by 2; a/2. So 95% confidence would have an alpha of 5% or .05. That is divided evenly between both tails. So .025 probability in the lower tail and .025 in the upper tail. If alpha was 10% or .10 then a/2 is .05 probability in the lower tail and .05 probability in the upper tail (90% in the middle).
in example1 it was mentioned that sample mean was 4 but you considered it as sample std deviation while working on the solution.
This video and all of your videos are absolutely fantastic, you're saving me in Statistics!!
Brandon, your videos have been true lifesavers for me! Currently enduring BUS308 Statistics for Managers, and the explanations you post ring like lead crystal: I reference each video I use in the assigned discussion forums. Spreading the word!! Thanks again :)
Hi Cherie! Problem #2 was adapted from a textbook I use in my work. So the standard deviation comes from a similar problem. I assume the problem was part of a study of some sort at one time.
i really like your word of encourage, thumbs up.
+Ariel T You are very welcome. To do it we have to believe we can do it.
Brandon, your videos are a great help to me. You explain everything perfectly. Thank you so much.
Hello! Yes you are correct that is a typo. :( I will place that correction in the description and as an annotation. Also, if you go to my channel page, all my videos are in playlists in their natural order. Each playlist follows the same order a textbook would and inside each playlist the topics are also in order. Thank you so much! - B
Brandon Thank you so much for ALL the videos. Much gratitude .Thank you
In 00:58 , it is saying that the sample mean was 4$. But you wrote sample standart deviation s=4. Is there a mistake in the question?
plz answer this
Yes, it must be a mistake! 4 is the standard deviation of the sample. Very confusing though.
It's got to be standard deviation, coz a mean of $4 would be very sad for a restaurant!
Thanks for always teaching the concepts... because you make me want to learn more about stats...I am enjoying my journey...
Great videos and thank you. I think I found an error in Statistics 101: Confidence interval problems, sigma unknown. In the example of the false customers when the sample size was changed from 30 to 50, the t statistic remained at 2.045; it should have been closer to 2.0.
Your explanation is way better than Udacity's lessons. Thank you so much.
Regarding Problem #2: I am stumped on how you got an "xbar" of 3661.5. Where did this come from? I have tried so many formulas and can't figure this out. Thank you for your videos. Very helpful!
Hi! Pretty sure that is just a given value in the problem :)
Hi Brandon, much appreciated! In Holland 2.045 x 276.9 = 566.26. Which means that the subsequent calcs need adaptation as well
Hi Brandon.... for example 1 qn says sample mean is 4 which means x-bar is 4 but you wrote sample std deviation as 4. Can you clarify this
I understand everything here except how you came up with your t scores when looking at the concern points. For example, when moving the sample to 50 people, how did you derive the t=2.38. i imagine it's a simple answer.
Great lecture. Very clearly explanation, very helpful. Thanks so much!
thank you so much. but just a question. in the last example you increased the sample size but you did not change the t statistics (2,045) even if we have a new degree of freedom ??? please answer me if you get my comment
How do you come up with the sample standard deviation 1958 seconds?
I have been going insane trying to figure this out! Also, where did the 4320s go? Wouldn't that be n? Someone, anyone HELP please!
Dr.Brandon , When you changed the Sample Size to 50 instead of 30 , the degrees of freedom changes alspo to 49 and so 2.045 should change to 2.0 instead, am i right?
Sorry Dr.Brandon , i didn't see the sticky note on the video , yes you have corrected that , thank you , now everything makes sense .
Brandon, Thank you SO much for doing these videos. They are - hands down - the most clear and easy to understand a all I have come across. I do have a question. In your example in this video on False Customers, where you change the sample size from 30 to 50, why do you only calculate a new estimate for the standard deviation and notalso determine a new t table value based on 49 degrees of freedom? Thank you in advance.
How did you calculate the t score value for 4320 seconds in the example where the df is 29?
Hi Brandon,
One question here. If the standard error of the mean is the standard deviation of the sampling distribution then in the case where we do not know the population standard deviation why don't we calculate the standard error of the mean by calculating the standard deviation of the sampling distribution??
Thanks.
I am having problems with The scores on an examination in finance are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination.
421, 422, 468, 507, 530, 567, 586
Send data
to Excel
Find a
90
%confidence interval for the population standard deviation. Then complete the table below.
Thank you. My professor mentions 4 types of Confidence Intervals. I am not sure what they are because textbook is not clear. Also need to know the formula for each one - again one runs into the other and I am unable to distinguish them.
Thank you for these videos. I'm reviewing and I'm at a point of confusion. Regarding the example of the show rooming, is this an example of a hypothesis test that will be explained in further videos? Or, is the example not related to hypothesis testing?
You really nailed it, awesome explanation. Now a burning question which I am sure everyone has, how do you obtain and calculate the SAMPLE STANDARD DEVIATION?
Nevermind a light bulb clicked! I can calculate the variance and then squareroot that to find the standard deviation! Like if I am right
Would you mind simplifying your thought process out? My light bulb isn't working!
@@denissedeo895 Just like the way you calculate Mean of a Sample, calculate the Variance of a Sample and find the Square Root of that Variance.
Hope this shed some light to help you finding the Root.
How do you come up with the sample standard deviation 1958 seconds?
I have been going insane trying to figure this out! Also, where did the 4320s go? Wouldn't that be n? Someone, anyone HELP please!
1st question : where do 1958 population standard deviation came from?
remember, you collect a sample with the size of 30 (30 different people with different bills).
from this 30 measurements, you can regularly calculate mean (3661.5 $), variance (sigma(xi-3661.5)^2) divided by 30-1), and standard deviation (square root of variance).
you've got to calculate each of 30 samples then substract with the mean, etc. to get the standard deviation. to cut on time, Mr. Brandon merely provides us with the standard deviation first hand since he assumes we all know how to calculate it (see playlist 02 on Mr. Brandon's channel)
2nd question : i don't get exactly what you mean, but 4,320 seconds is just assumption, so the manager of this store makes an assumption, -- *if more than 15% of the worker's average time (4,320 seconds) is spent on false customer* --, then we got a problem.
the first samples shows a range of average time spent on customer of 2930-4392 seconds >> so, from this ONE sample, the manager concludes that they got a problem with false customer
the second samples (which is larger in size, and therefore more representative) shows a range of average time spent on customer of 3,104 to 4,218 seconds >> notice that the range of average time spent on false customer doesn't reach the 4,320 marks, therefore they've got no problem with the false customer
In the initial slide of example, sample mean is written as $4, instead of sample standard deviation. Next slide is correct. Please rectify/ revise initial one.
thanks so much for these videos i can now face bio statistics staff with confidence thanks again Brandon
Hi
How did you get the t=1.84 and t=2.38? Thank you for the awesome videos
Its the concern point (4320) minus the sample mean (3661.5) divided by sample standard deviation (357.48). (4320-3661.5)/357.48
I was very confused too, it took me a minute to figure it out.
These videos have helped so much. I've shared them with others in class.
I do have a question about CI that still isn't clear. Why with a binomial distribution do you square root pq/n, when for these problems you only root the n?
How did you get the x-bar of 3661.5 in the show-rooming problem?
Hi! It is just a "given" in the problem. :)
@@BrandonFoltz ah.. I see. Thanks!
@@BrandonFoltz it would be nice to put the x-bar value in the slide as well, so that your viewers would be less confused on small things like this. Overall, your videos are great and super helpful for a struggling student like me. Kudos to you!
Great video!
Question: the standard deviation formula is different depending on whether the data is the full population or just a sample.
When calculating the standard deviation for a sample to then use in the SEM formula, presumably we should use the sample std dev formula.
Is that correct? Or do we say that perhaps since the t-distribution we're going to compare against is already using degrees-of-freedom, perhaps we only do it once?
One caveat to the customers draining resources example. When you increased the sample size from 30 to 50 and held everything constant, that's not to say that the extra 20 observations would have in fact changed the sample mean an thus the sample standard deviation and thus the standard error of the sample and therefore the threshold could have fallen into our confidence interval, right?
Hello! Can anyone please guide how he come up with t value of 1.84 @ 11:06? and t value of 2.38 @ 15:01?
you can find the t-value by looking at the t-table or using an excel function. if you're using excel use the t.inv(prob;df) function to find the points.
i'll go over the basic a bit here,
-why do you need to find t-value? to find the lower and upper point where 95% of the sample distribution falls
-what do you need to find t-value? just the fact of how many % of confidence interval (90%, 95%, 97%, etc) do you want to have, and what are your degrees of freedom (size of sample minus 1)
-how do you find t-value? on the t-table, just find the degrees of freedom of 29 on the row, and 5% significance on the column, you'll find 2.042, which is exactly the same as the video.
on excel, you use the function =t.inv(probability; df) >> =t.inv(probability0.975;29) we use 0.975 (97.5%) because from the left-most (-infinity), to lower point (2.5%), to upper point(97.5%) contains 97.5% probability.
What is the difference between saying that there is 95% probability that the population mean falls whithin an interval and saying that we are 95% confident that this interval contains the population mean?
Hi, I was hoping to find an explanation on how to find the t value for the concern point. so far I know only to use degrees of freedom and the alpha / 2. how do we know the df for the concern point. or if we know the concern points value along the x axis how do we work t point backwards. is this in another video? I don't even know how to search internet for this instance. if you see this question I'd love an answer. Thanks!
Hi Brandon, let’s assume I want to calibrate a lab balance using a reference weight. I measure 30 times the weight and calculate the mean and standard deviation. So I can estimate a confidence interval for the sample mean. After some time I want to check the balance to verify it is still accurate. I take a double (or triple) measurement of the reference weight and obtain an average value. How do I know if the value obtained is within the acceptable range?
awesome video! thanks! do u have one for proportions? great job-thanks again
Hi Brandon. Thanks heaps for doing this awesome work. I am learning a great deal out of your videos. Just a question.. in the 2nd example when the mean was increased from 30 to 50.. the t distribution through the chart was still 2.045 (which I reckon based on 29) Shouldn't we put 1.96 when the sample size is 50? Thanks
if you change your sample size n 30 to 50, the sample mean will be change also?
+CID amber suppose it remains the same
Hi, Great video and explanation. I have small question regarding the example 2 customer service drain, normally when you try to find t value, we use degrees of freedom and alpha/2 = 0.025 but when you found that the concerned point is coming inside the calculated confidence interval,you found the t value as 1.80 at 4320 seconds. How did you get that t value? did you change the alpha value? please assist with your answer. cheers :)
So the difference between the "standard deviation of the sample" and the "standard deviation of the sample means, ie the standard error of the sample" Is that the former tells you the standard deviation of ONE sample, while the later tells you the standard error of the distribution of multiple sample means if you were to take multiple samples?
hello,
we find t value based on alpha value at 95 % confidence interval and degrees of freedom. what if i need t value for 93.6 % confidence interval? it is not shown in t distribution table. then how to find it? and what if i know some specific value of sample beyond 95% confidence may be at 96.5 %. how to find t value there? please answer :)
Brandon, I am having major problems with confidence levels. In one example (looking at 95% confidence level) I find that I need alpha divided by 2 (.05) in order to find z score and on another example I find that the confidence level - of 95% has to be divided by 2 to find the z score. I am confused.
Just heard you mention alpha at the beginning of this example. Am I correct to assume that the T distribution table refers to the equivalent of alpha. What does alpha has to do with the t distribution? Maybe that is what I am missing. Thank you.
Hi Brandon, thank you so much for your videos. i have a question though. the standard deviation is 4 (on video 2.23secs) so you put this into an equation and get a standard deviation of 1.03. whats the difference between the sd 4 and the sd 1.03? would be really gratefull if you could come back to me asap. thanks again!
I have done the example but still can't come to the answer
is the t-table the same as the z-table? in my class we have not learned anything about a t-table?
I want to ask.. interval from 22.21 to 4 and from 17.79 to -4 is not the margin error ? how we can naming this 2,5% from left and right ?
And oh, what is the name of the textbook that you use
I think there's an error. @1:00 xbar is mentioned instead of s @1:05 s is assumed. I guess it was a typo, mean instead of sample. I love your videos. It would be cool if a playlist was made with the natural order.
in both examples - Why is the sample standard deviation not substituted directly as sigma (X bar)in confidence interval formula ( X bar + 2.145sigma (x bar)) ? The sample standard deviation is divided by square root of Sample size like it was done in population std deviation problem...Why is it not substituted in formula directly...Really confused..Can someone please clarify?
Hi Brandon: How do you know when both tails are being used?
This video helps me a lot!
Hello
Brandon Foltz
This is my first time to go public and ask a simple question?
Where do you get X=3661.5
* Is the number random select or a sample number from you?
* OR given number that you select?
I am just stuck with the X=3661.5
Probability you had shown how to get it but I need to be refresh up
or give me some clues
or tell me to go to this video and there is where the x=3661.5
Okay
Thanks
Jose Sanchez Moreno Hello and thanks for watching! It is just a "given" in the problem like you may see in a textbook. Hang in there!
Quick Question: Is this a mistake the first problem in the wording says that the sample mean is $4, but the next slide it was used as a standard deviation. Is that an error or am I under looking something? Love your videos by the way
Excellent 5 🌟
Very good video
from where you got x bar=3661.5???
YOU ARE AWESOME!!!!!!!!!!!
Where does 2.21 comes from?
it comes from T value times the Sx
11:20 i didn't understand how you got the value of t
Brandon for the POTUS.. Who needs Trump?.. Excellent video and your explanation technique!!
15:00 why Tvalue didn't change? Our n changed and I think Tval is dependent on n just like SE.
great video
how did you get s=4?
Great! Thanks!!
Thanks
sorry i watched the video again... 1.03 is the standard error of the mean! face palm.
Oh sorry.. disregard this. I didn't read the correction bubble :)
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