Standard Error (of the sample mean) | Sampling | Confidence Intervals | Proportions

0:00 Introduction
1:20 Definition
6:40 Confidence Intervals
12:51 Proportions
17:16 Challenge Question

Series music by Purdy.
Song: 3 Friends to the Stars

[abhishekdileep]

Seriously man, i can't believe how much of a difference creators like him make. Honestly they are the reason why people like us have honest confidence to accomplish things in the field that we are doing ! you are the reason growth of this societies happen

[joel-uni-acc]

"so heres the formula.. now don´t get too excited."
thanks justin
the impact your videos are having on all these peoples futures is inspiring

[TheSouzaMarvin]

SE(x) = s/√n
s/√n+x = ½( s/√n)
√n+x = 2√n
n+x = 4n
x=3n
x=3*20=60

You will likely require 60 more measurements, provided the standard deviation remains the same when the sample size increases.

[goodfornothing]

Hey Brother,
You are the best statistics tutor on the YouTube.
Thank you so much.
Appreciate your good work.

[Anukul]

(S/root(n))/(S/root(a)) = 2
a/n = 4
a = 4n
if n = 20, a = 80, so 60 more observations.

[peterbuchanan]

The delivery at 2:06 was absolutely perfect and had me chuckling for a good minute

[mohemmedansari]

Very very helpful. I landed here in my journey on Lean - 6 Sigma education.This series and the other videos from you helped me sharpen my statistics concepts.

[robhemp]

12:19 shows the 500 count interval below the sample mean.

[swarnachoppella]

Challenge: S.E(sample mean)/2=sample S.D/√(4n)
so it will be 80 measurements (4*20)
So we need 60 more measurements

🙂thank you for sharing wonderful lectures

[igsor]

Thanks for a great stats course!
timestamp 12:45 - Why sample mean 114.7 is out of a confidence interval [110.8, 113.1] ?

[j]

EDIT: this answer is wrong, but in an interesting way. So I’m going to leave it here, with a correction posted below. I have posted a separate answer with what I believe is the correct math for the challenge question.

If you increase the sample size, two things will happen: (1) n will increase, and (2) you’ll be adding random data points, so they will probably end up with a somewhat different mean and standard deviation.

About #1: n appears in the denominator of the formulas for standard deviation *and* SE, so an increase in n is sort of double counted; combining the formulas, you get a denominator of [sqrt(n-1)*sqrt(n)], which is pretty close to just n, especially for larger n. So in short, doubling n should get you close to halving SE.

About #2: The change in mean doesn’t affect SE, but if the variation about the mean changes as new data comes in, it would alter the numerator of standard deviation, which in turn alters the numerator of SE. It’s beyond my skills to know if this is likely to change up or down as n grows, so I’ll just assume that, on average, actual variation about the mean does not change. Thus, all that matters is the increase in n, where a doubling of the sample would approximately half the SE.

[armaosk]

Hi Justin, I've made it but I cannot find the videos of outliers, boutique measures and Range+IQR) which, according to the diagram should be on the list. I searched your channel but I could not find them. Are they posted yet?

[shreddersengupta]

( SE/(ROOT(20 + x) ) = SE/2 => x = 60, therefore 60 more. Therefore, Total 80 will make Half the SE

[bookender]

Is it possible for the confidence intervals not to contain the sample mean, as shown at 12:39 for n=500? If I understand this correctly, it is saying that when you have a sample mean of 114.7 there is a high likelihood that the population mean is between 110.8 and 113.1. Unless we had other information about the population, this doesn't seem reasonable to me.

[dakshpatel]

Very Informative. I recommend this channel for the students who want to learn statistics for Data science and Machine Learning. As almost all concepts are covered that essential to learn for understanding the Machine learning models...
Thank you and really appreciate your work

[AndrewClark]

@zedstatistics Perhaps, I've missed something, but If the confidence intervals are the sample mean +/- the (SE * t score), how is the 95% confidence interval score for n=500 *lower* than the sample mean? Sample mean is 114.7, and the 95% confidence upper threshold is 113.1 in the slide (12:46). Should it not be 0.55 (SE) * 1.964729 (T score) + 114.7 = 115.78?

In the slide, we have n = 5 with an even difference +-/ from the sample mean (15.8)
with n = 50 we have (6.9 below and 0.3 above)
with n = 500 we have (3.9 below and 1.6 below)

[carly]

Loved it! I am new to statistics and ML and this has been extremely helpful in understanding the basic concepts 🙏

[jennifershao]

80 measurements but on the condition that the standard deviation remains the same with the observations increasing from 20 to 80.

[wuyunhua]

did you make some mistakens when compute confidential interval as n=50 and n=500, for example, as n=50, and x bar equals 115.3, your confidential interval is [108.5, 115.6], however, (108.5+115.6)/2 not equals 115.3. looking forward your help, thx

[shnks]

i dont think the underlying distribution needs to be a normal random variable...the sample mean distribution will always follow a normal distribution due to central limit theorem...also z statistic is used in case population variance is known otherwise t statistic is used