Thank you so much for posting this video. I am doing all online classes and got stumped on how to find the p-value with a - t score. This helped me so much!! Thanks!!
Your videos are so helpful! Also taking Elem. Stats online, your review w/the MathXL questions has saved me! I have suggested that my prof either make some of her own or borrow yours for the "help" folder. THANK YOU!
Thank you! You made it clear and easy to solve this problem. I wish my Park Instructor can do a video and teach us how to do in the University Page. I also like the idea that you walk us through using the TI-83 Plus Calculator. Appreciate your explanation...
Omg!!! Thanks so much! You are a life saver! Could not understand how to use my calculator to input these problems. Statistic is going to give me grey hair! Lol. But seriously thanks again!!!!!!!!!!’
Thanx tons my friend. I kept using Normalcdf and realized something wasn't right as my professor did not give us such example considering the t-distribution given.
Is there any way I could use this for this specific problem? "In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%." Where would I place the 9999? because its a 2 tailed test, so I don't know which side to put it on.
If you are looking for the P-value for a left or right-tailed test, is the process the same (-9999 for negative t-statistic, 99999 for positive t-statistic)? With the only difference being that you do not multiply the result by 2? If you are given a right-tailed test with a negative t-statistic, would you use positive 99999? Or is a negative t-statistic always associated with -99999 regardless of the type of test.
Yes that's correct. -999999, t statistic, df for the left tailed test and t statistic, 999999, df for a right-tailed test. The sign on the test statistic doesn't matter. If you did a right-tailed test with a negative t value you will get a p-value > 0.5, which just means you won't reject H0. It also means you probably should have done a left-tailed test, but there's nothing "wrong" with doing a R-T T with a negative t nor a L-T T with a positive t. And as you mentioned, just don't multiply by 2 at the end.
TakeRenaHome i believe it depends on the test statistic as well, if it is negative, the lower value is -99999, but if the test statistic is positive then the lower value is the test statistic and the upper is 99999.
So I eventually got to what I needed to learn from your video so I gave it a thumbs up. However, some constructive criticism, I had to watch this at 1.75 speed you went through it very slowly.... also, as i'm sure you've read over and over the volume is terrible. Thank you for your help though =]
What was your test statistic? Was it a t, z, or possibly a different type of test? Was the standard deviation known? Was this about a mean, proportion or some other statistic?
Heres the problem: Many people believe that the average number of Facebook friends is 140. The population standard deviation is 38.4. A random sample of 47 students in a particular county revealed that the average number of Facebook friends was 145. At a= 0.05, is there sufficient evidence to conclude that the mean number of friends is greater than 140? Use the p-value method. Use a graphing calculator. The p-value is 0.1860... But my calculator shows .37203....
+Shon9tilR this video is for a TWO-tailed test. In the problem you are working on the test is to determine if the average is "greater than" the old average. That means this is a ONE-tailed test (right tailed) and you do not need to double the p value. Notice what you wrote is double (with some rounding) the correct p-value. "Greater than" or "less than" is the wording in one tailed tests. "Not equal" or "different" are typical phrases suggesting a 2-tailed test.
+Shon9tilR Did you run a test in the STAT, Test menu? If so, if you selected =/= (not equal) it performs a 2-tailed test. That is the default setting. But if you need a 1-tailed test you would need to switch to > (greater than) or < (less than)
Is there any way I could use this for this specific problem? "In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%." Where would I place the 9999? because its a 2 tailed test, so I don't know which side to put it on.
Thanks! This is about the fourth or fifth video I have watched claiming to explain this process, but the first one that actually explains it.
Love you for posting this. I hope you know you're a great person and deserve all the praise you have gotten in your life
You, sir, are a life savior. Thank you
Thank you so much for posting this video. I am doing all online classes and got stumped on how to find the p-value with a - t score. This helped me so much!! Thanks!!
watching these before my test this afternoon is really helping me understand, thank you so much
Your videos are so helpful! Also taking Elem. Stats online, your review w/the MathXL questions has saved me! I have suggested that my prof either make some of her own or borrow yours for the "help" folder. THANK YOU!
Thank you so much for posting this!!! I have been dealing with a similar problem for way too long! Thank you!!!
I don't think you understand how much this helped me!
Exactly what I needed, thanks!
Thank you for this video!!! I am taking STATS online and this was a huge help!
You have managed to explain this better than my old ass professor who's probably been explaining this poorly for hundreds of years. Thank you so much!
Thank you! You made it clear and easy to solve this problem. I wish my Park Instructor can do a video and teach us how to do in the University Page. I also like the idea that you walk us through using the TI-83 Plus Calculator. Appreciate your explanation...
OMG Thank you!!!!! Most helpful video on youtube lol. I swear youre a life saver!!!
Thank you so much!!! This saves me so much time and frustration.
straight to the point!!! good job! thanks
THANK YOU for covering the negative T distribution.
I appreciate you making this video.
You just saved me a lot of frustration... Thanks a bunch...
You are a life saver 🙌🏼
Thank you for posting this! None of the other examples show how to find the p-value with the T data
Thank you so much for this in 2021!!!!!
Life saver! Thank you!!
helped me tremendously, thank you so much!
YEEEESSS, Thank you so much!
Is it just me or is this video extremely quiet? I can't hear a thing.
Wow you are awesome!!! THANK YOU
thank you thank you
this really helped a lot
i was struggling with finding p value last 2 hours LOL
Same here! LOL!!!!!!
Thanks man. Helped me more than my book or teacher
Omg!!! Thanks so much! You are a life saver! Could not understand how to use my calculator to input these problems. Statistic is going to give me grey hair! Lol. But seriously thanks again!!!!!!!!!!’
Super helpful !
Thanks for the vid! Very helpful
Couldn't really hear it, but this is exactly what I needed. Thank you!
Sorry for the audio. I bought a better headset because of the comments on this video...and because the old one was terrible. :)
Thank you so much!
Thank you very much!!!
Thanx tons my friend. I kept using Normalcdf and realized something wasn't right as my professor did not give us such example considering the t-distribution given.
OMG! THANK YOU!
This is very helpful! I only wish the audio were a bit louder. But it was very informative!
Thank you!!!
Omg thank you so much, your video helped a lot life saver thank thank thank you so much
You saved me ,, thank you so much.
Thank you!
thank you so so so much you just helped me so much
Awesome thank you!
I wish my professor explained this like you did!
this helped a bunch
So do I. But since I have to take it I am thankful for videos like this one.
You saved my life
THANK YOU
Bless up🙏🙏🙏
He has such a nice voice.
Thank you. I was seconds away from breaking my calculator....
thanks mister
this helped me.
thank you!!!!!!!!!!!!!!!
thank you
The original P value is for one-tailed tests. You have to multiply it by 2 for two-tailed tests.
very helpful
thnx mr dave
I tip my hat to you sir! Teacher never explained how to get the P-value for non-TI-84 calculators.
Yes, as he said, the initial answer is for the one tailed test. If you want the answer for a two tailed test, just multiply the answer by two.
thanks alot
Yes. Just don't multiply by 2 at the end.
Thank you
Is there any way I could use this for this specific problem?
"In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%." Where would I place the 9999? because its a 2 tailed test, so I don't know which side to put it on.
thanks.
thanks
Make sure you're using tCdf and not tPdf. Can you let me know what you put in?
If you are looking for the P-value for a left or right-tailed test, is the process the same (-9999 for negative t-statistic, 99999 for positive t-statistic)? With the only difference being that you do not multiply the result by 2? If you are given a right-tailed test with a negative t-statistic, would you use positive 99999? Or is a negative t-statistic always associated with -99999 regardless of the type of test.
Yes that's correct. -999999, t statistic, df for the left tailed test and t statistic, 999999, df for a right-tailed test. The sign on the test statistic doesn't matter. If you did a right-tailed test with a negative t value you will get a p-value > 0.5, which just means you won't reject H0. It also means you probably should have done a left-tailed test, but there's nothing "wrong" with doing a R-T T with a negative t nor a L-T T with a positive t. And as you mentioned, just don't multiply by 2 at the end.
Thank you very much! This video was incredibly helpful!
what if it's a one-tailed test? do we have to use another number besides 9999?
Thanks
TakeRenaHome i believe it depends on the test statistic as well, if it is negative, the lower value is -99999, but if the test statistic is positive then the lower value is the test statistic and the upper is 99999.
thnks
So I eventually got to what I needed to learn from your video so I gave it a thumbs up. However, some constructive criticism, I had to watch this at 1.75 speed you went through it very slowly.... also, as i'm sure you've read over and over the volume is terrible. Thank you for your help though =]
does this work for one tailed tests?
my calculator said err syntax how can i fix that ????
I followed an example ik my book but my p value is not coming up the same. What can I do?
What was your test statistic? Was it a t, z, or possibly a different type of test? Was the standard deviation known? Was this about a mean, proportion or some other statistic?
Heres the problem:
Many people believe that the average number of Facebook friends is 140. The population standard deviation is 38.4. A random sample of 47 students in a particular county revealed that the average number of Facebook friends was 145. At a= 0.05, is there sufficient evidence to conclude that the mean number of friends is greater than 140? Use the p-value method. Use a graphing calculator.
The p-value is 0.1860... But my calculator shows .37203....
+Shon9tilR this video is for a TWO-tailed test. In the problem you are working on the test is to determine if the average is "greater than" the old average. That means this is a ONE-tailed test (right tailed) and you do not need to double the p value. Notice what you wrote is double (with some rounding) the correct p-value.
"Greater than" or "less than" is the wording in one tailed tests. "Not equal" or "different" are typical phrases suggesting a 2-tailed test.
+Shon9til 0.1860x2= .37203. Ok! I wonder why the calculator doubled it?
+Shon9tilR Did you run a test in the STAT, Test menu? If so, if you selected =/= (not equal) it performs a 2-tailed test. That is the default setting. But if you need a 1-tailed test you would need to switch to > (greater than) or < (less than)
Watch the video.
ti-83 dont have the invt(p,df) feature.
Try Desmos. It is free. There’s a bunch of videos on CZcams
Fix the audio.
Thank you!
THANK YOU
He has such a nice voice.
Is there any way I could use this for this specific problem?
"In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%." Where would I place the 9999? because its a 2 tailed test, so I don't know which side to put it on.
Thank you!
Thank you!