Thank you for actually showing the Z table, to me that is the most confusing part. The z table I have is only for the right half of the curve and gives Z values to one decimal place. Helpful video!
This is a very smart explanation. I am really lighten up with the understanding of z-test two tail test for 1 sample for mean. Thumbs up. wwwooooowooo!!!!!!!!
God bless you. I've watched SO MANY videos in the last couple of hours about this exact topic and this is the first one that I was able to understand and apply it to my homework. I love you king
Ah yes of course thanks! Sorry didn't see Type I and Type II errors there. That makes a lot more sense. Thanks again for your lectures. I am a distance learner and would not know what to do if I only had my textbook. Really appreciate these.
For those who confused why we use significance level is 0.05%. A 5% significance level strikes a balance between the risk of Type I errors (rejecting a true null hypothesis) and Type II errors (failing to reject a false null hypothesis). It is a common compromise that is considered reasonable in many scientific disciplines. So it is mostly used if significance level is not given in question.
dude I'm brazilian and my teacher couldn't do in my native language what u've done in a foreign one . I don't know if you're really good of if she's really bad. anyway, thank you so much
You reject null hypothesis in both cases if the results are greater or lower than critical value...infact we are fail to reject null hypothesis when the z value is greater than critical value..beacause this way it doesn't come into rejection reigon.we will only reject when value is lower.
Actually there is one confusing thing, as you would expect that 140 lays between 2 and 3 sigmas and your z is not 14.6, but (140-100)/15. Yet formula tells you to divide on sigma/sqrt(n). The answer is that IQ itself is not normally distributed, the Mu of IQ has normal distribution, and sigma of average is not 15. So the chart in point 3.) is not correct. Correct chart is that you build normal distribution around average number of 140 which has std = 15/sqrt(30) and then you see that Mu=100 lays within 14.6 errors comparing to the statistic 140
Hi there.. these are very helpful thank you so much. I really am lost when you speak about alpha though. Which lecture do you deal with the concept of alpha? I have watched them all but could not find anything? Thanks again.
how did you decide the alpha to be 0.05? why isn't it like 0.1 or sth like that? and then why do you conceive 2.5% to be enough to reject the null hypothesis? what kind of measurement did you have?
Since it is a 2 tail-test, therefore u have to divide your level of significance that is alpha's value by 2. Hence, if 0.05/2 we get 0.025. now in table to get our z-test value we have to look for 1-0.025 which equals to 0.9750 or 97.5% . I hope my explanation was helpful for your doubt.. though it's more than 7years from now.. 🙈🙈
@@priyamthapa7172 But the body is still 95% of the curve because you have 2 tails of 2.5%. If the body was 97.5% wouldn't that imply a 1-tail test with an alpha level of 0.025?
@Whitney D , Lets take it this way, 0.9500/2 = 0.4750, there are different types of z tables if you look on z table that is area between 0 and z. 0.4750 falls on 1.96
So the sample means are normally distributed around the population mean, and the standard error is used to estimate the standard deviation of the distribution of the sample means. Hope that helps 3 years later!
It is basically OK. One thing bothers me though. The prob density plot in point 3 shows the population IQ. But the decision we are going to take about the hypothesis refers to sample IQ which std dev is smaller than that of the whole population. So the values at 2,5% should be approx. 94 and 106 as with this prob. density the result of 140 IQ is effectively compared. Or maybe I am wrong?
@@arctic219 i think this requirement is a mistake. If you know the standard deviation, you do not have to make the assumption about the n>30. If it is not known, then it will be a t distribution. Z-distribution is already a normal one, and it is valid even for one sample. The requirement for n>30 would be valid if the underlying population distribution was not originally normal.
No, you only use a t-test when you either don't know the population standard deviation and/or the sample size of the data is really small. In this example, he was given the population standard deviation (which was 15) and the sample size was large enough, hence why he used a z-test and not a t-test.
Thank you!! But just FYI the way you talk in the video is hard to understand. Like you're saying the words too fast and it slurs together. Just for future videos, you might want to know that!
so with an alpha of 0.05 (5%) we divide that by 2 since it's 2 tailed which results in 2.5%. We take 100% - 2.5% which gives us 97.5%. We then divide that by 100 to transform it into a decimal, resulting in 0.9750 which corresponds to a z-score of 1.96. Hope that helps :/
he used percentile tables; use a z-table showing the right tail (be sure to double your p-value from this table if you're doing a 2-tailed test like here
It's out of 100, 95% falls in range while 5% falls out of range. Half of the 5% fall on opposite ends of the spectrum. 2.5% fall on the left side of the tail, and 2.5% fall on the right side of the tail. For example, if they found the results of the medication had no effect(left side) or a large effect(right side). Again, sine it is out of 100, you subtract the 2.5 to get 97.5. Then you look up the 97.5 in a "z score table" to find the z score, which is 1.96. Make sure you sure the correct z score table.
I'm assuming this is a right tail test as the sample mean is on the right side compared to the population mean, with respect to the alternative hypothesis if we want to reject or accept it we have to find the z score so we only consider the rejection region to be above the 97.5% region.
1.If a null hypothesis can be rejected at level alpha=0.03, what can u say when alpha=.05 2.what can u say when alpha =0.01 Dude this question has been bugging me all week ..i really will appreciate if u will be kind enough to explain this.
Thank you for actually showing the Z table, to me that is the most confusing part. The z table I have is only for the right half of the curve and gives Z values to one decimal place. Helpful video!
This is a very smart explanation. I am really lighten up with the understanding of z-test two tail test for 1 sample for mean. Thumbs up. wwwooooowooo!!!!!!!!
God bless you. I've watched SO MANY videos in the last couple of hours about this exact topic and this is the first one that I was able to understand and apply it to my homework. I love you king
very well explained and allowed concept to sink in. thank you.
Still relevant 12 years later.
Ah yes of course thanks! Sorry didn't see Type I and Type II errors there. That makes a lot more sense.
Thanks again for your lectures. I am a distance learner and would not know what to do if I only had my textbook. Really appreciate these.
For those who confused why we use significance level is 0.05%. A 5% significance level strikes a balance between the risk of Type I errors (rejecting a true null hypothesis) and Type II errors (failing to reject a false null hypothesis). It is a common compromise that is considered reasonable in many scientific disciplines. So it is mostly used if significance level is not given in question.
Very simple method of explaining a complex problem
Thank you, very clear and easy to understand. Much better than my textbook!
J West by
Thank you so much for this video! had a hard time on how to calculate the table value. Thanks to you, finally done with my stat report
this is the video i am searching for..... excellent explanation
@ statslectures: Very nicely explained. Thank you!
Very clear and so useful, thank you very much!!!!
dude I'm brazilian and my teacher couldn't do in my native language what u've done in a foreign one . I don't know if you're really good of if she's really bad. anyway, thank you so much
Great explained! thanks!
How did the 1.96 came?and the area in the body too
Very nice explanation, thanks a lot :)
i have 10 questions in my survey, does it mean i have to do this computation in each number?
where to get the z-score table?
very well Explained
clearly explanation. Thanks!!
Thanks very much. It is very helpful
You’re a lifesaver!
Why was there no probability statement?
You reject null hypothesis in both cases if the results are greater or lower than critical value...infact we are fail to reject null hypothesis when the z value is greater than critical value..beacause this way it doesn't come into rejection reigon.we will only reject when value is lower.
Actually there is one confusing thing, as you would expect that 140 lays between 2 and 3 sigmas and your z is not 14.6, but (140-100)/15. Yet formula tells you to divide on sigma/sqrt(n).
The answer is that IQ itself is not normally distributed, the Mu of IQ has normal distribution, and sigma of average is not 15. So the chart in point 3.) is not correct. Correct chart is that you build normal distribution around average number of 140 which has std = 15/sqrt(30) and then you see that Mu=100 lays within 14.6 errors comparing to the statistic 140
Thank you so much Sir for this video.
But small doubt in z test we take large sample problems but here n is 30 that means this is a small sample text
Where didi the 0.05 came from
Can anyone tell me how we got to know the area of the body not able to understand kindly help
Why is it two tailed because comparing higher or lower iq, I thought that 2 tailed was comparing difference while 1 tail is comparing higher or lower
It was a nice tutorial, But i didn't understand that why instead of %2.5 of z you selected 1.96 for calculating the body percentage ?
+Hasibullah Sahibzada (%2.5+%2.5=%5 which is 0.05). A range of plus/minus 1.96
Ah, thank you, my friend.
Thanks
Good explanation
Thank you!
Why didn't we choose an alpha of 0.01 but 0.05?? Please anyone who understands to explain to me.
Thank you
Thanks for this! :))
Hi there.. these are very helpful thank you so much. I really am lost when you speak about alpha though. Which lecture do you deal with the concept of alpha? I have watched them all but could not find anything?
Thanks again.
Alpha is always 0.05 unlesss it states otherwise
THANK YOU SO MUCH OMG
how did you decide the alpha to be 0.05? why isn't it like 0.1 or sth like that? and then why do you conceive 2.5% to be enough to reject the null hypothesis? what kind of measurement did you have?
it is most common to assume alpha to be 0.05 if it is not given
Hi This really helped me understand the material. However I still dont understand the explanation about why 97.5% was used instead of 95%.
Since it is a 2 tail-test, therefore u have to divide your level of significance that is alpha's value by 2. Hence, if 0.05/2 we get 0.025. now in table to get our z-test value we have to look for 1-0.025 which equals to 0.9750 or 97.5% .
I hope my explanation was helpful for your doubt.. though it's more than 7years from now.. 🙈🙈
@@priyamthapa7172 But the body is still 95% of the curve because you have 2 tails of 2.5%. If the body was 97.5% wouldn't that imply a 1-tail test with an alpha level of 0.025?
@Whitney D , Lets take it this way, 0.9500/2 = 0.4750, there are different types of z tables if you look on z table that is area between 0 and z. 0.4750 falls on 1.96
its very useful. thanks..
Can someone explain why you need to use the standard error, and not the given standard deviation of 15, as the denominator for the formula of z?
So the sample means are normally distributed around the population mean, and the standard error is used to estimate the standard deviation of the distribution of the sample means. Hope that helps 3 years later!
you are great
This explaination is very clear about the z test
Thank you
If the area in the tail is 2.5 percent how the heck do we get to 1.96 please help
Amazing.
how did you decide confidence coefficient to be 95%?
You mean the confidence level? And the reason why he put it at 95% is cuz he put alpha at 5%.
Where the probability has come from?
It is basically OK. One thing bothers me though. The prob density plot in point 3 shows the population IQ. But the decision we are going to take about the hypothesis refers to sample IQ which std dev is smaller than that of the whole population. So the values at 2,5% should be approx. 94 and 106 as with this prob. density the result of 140 IQ is effectively compared. Or maybe I am wrong?
In the next video same n value but you have used t test .
Then how can we recognise whether it is a t- test or z-test problem
t test is for sample means, z test is for proportions
its good dude
why is a requirement to have number of datapoints greater or equal to 30 since we know the distribution?
its not required, but when n>30 the distribution is approx. normal because of the central limit theorem. if n
@@arctic219 i think this requirement is a mistake. If you know the standard deviation, you do not have to make the assumption about the n>30. If it is not known, then it will be a t distribution. Z-distribution is already a normal one, and it is valid even for one sample. The requirement for n>30 would be valid if the underlying population distribution was not originally normal.
How do you get 0.05%? I'm confused
He didn't "get" it from anywhere and it's 0.05/5%, not "0.05%." It's just the most common value that alpha is set to.
Shouldn’t this be a t-test since the sample is in means?
No, you only use a t-test when you either don't know the population standard deviation and/or the sample size of the data is really small. In this example, he was given the population standard deviation (which was 15) and the sample size was large enough, hence why he used a z-test and not a t-test.
Why is 0.975?? Shouldn't it be 0.95 since it's two tail?
"now im going to talk about one of the most basic" *breaksdown*
the area in the body is 95% so why did you choose 0.9750??? It is really confusing for me
This should help with your question - czcams.com/video/grodoLzThy4/video.html
bruce banner to the rescue!
Thank you!! But just FYI the way you talk in the video is hard to understand. Like you're saying the words too fast and it slurs together. Just for future videos, you might want to know that!
This video about z-test watch?v=TnTSrFhe0V8
Paige. No you have hearing problem s
Great video, didn't understand a thing
Then you oughta watch it again cuz it was real simple
you're probably missing some basic statistic's concepts
video starts at 1:48
How did he get 1.96?
so with an alpha of 0.05 (5%) we divide that by 2 since it's 2 tailed which results in 2.5%. We take 100% - 2.5% which gives us 97.5%. We then divide that by 100 to transform it into a decimal, resulting in 0.9750 which corresponds to a z-score of 1.96. Hope that helps :/
he used percentile tables; use a z-table showing the right tail (be sure to double your p-value from this table if you're doing a 2-tailed test like here
That's the same question I had. I'm reading through the comments & I still don't understand...
It's out of 100, 95% falls in range while 5% falls out of range. Half of the 5% fall on opposite ends of the spectrum. 2.5% fall on the left side of the tail, and 2.5% fall on the right side of the tail. For example, if they found the results of the medication had no effect(left side) or a large effect(right side). Again, sine it is out of 100, you subtract the 2.5 to get 97.5. Then you look up the 97.5 in a "z score table" to find the z score, which is 1.96. Make sure you sure the correct z score table.
I'm assuming this is a right tail test as the sample mean is on the right side compared to the population mean, with respect to the alternative hypothesis if we want to reject or accept it we have to find the z score so we only consider the rejection region to be above the 97.5% region.
1.If a null hypothesis can be rejected at level alpha=0.03, what can u say when alpha=.05
2.what can u say when alpha =0.01
Dude this question has been bugging me all week ..i really will appreciate if u will be kind enough to explain this.
wrong
Thank you!