@@kia2663 I think you could use any statistical software to compute those correlations. In R, two options are Hmisc:: rcorr(as.matrix(df)) and GGally:: ggpairs(df)
2:30 Aaaaaaand you've lost 95% of the audiences who were indeed looking for a "simple" explanation of this issue. Even assuming they are aware how to calculate corelation, how on earth do you expect them to know where you pulled all these figures out of?
Hello, can you please explain how you got .346? Which numbers did you use to find the mean. I got a mean of .446 instead of .336. Please help! :) Thank you!
+Jannette Reyes I think its the average of all correlation , except for correlation of the same item (equal to 1). (all no. in the table) hope it helped =)
I don't know whether this is the right way to calculate or not, but I got .346 by this way... (.35+.42+.25+.21+.31+.38+.36+.41+.46+.31)/10= 3.46/10= .346.
@@nirobkothopokothon Ah, yes! You don't include the 1's because that's the correlation value between a variable and itself. Most statistical packages show the 1's but they're not correlations between 2 different variables.
+Joseph Wehbe No. In such a case you should use polychoric correlations, which would give you ordinal Cronbach's alpha. SPSS doesn't do this analysis, unfortunately, but it can be done. Check out: Gadermann, A. M., Guhn, M., & Zumbo, B. D. (2012). Estimating ordinal reliability for Likert-type and ordinal item response data: A conceptual, empirical, and practical guide. Practical Assessment, Research & Evaluation,17(3), 1-13. In practice, people just run Pearson correlations with Likert data for Cronbach's alpha. Doing so underestimates reliability, but it's better than Spearman correlations, in this context.
@@novachrono2236 - no because even though the items nay be consistent, they could be consistently wrong as my professor said. You have to make sure they are measuring what you intended to measure.
What @Richii said. To make it clearer, there are 10 coefficients (all the floating point values, which are the values with decimal points, except for the '1.0' values). So, take the sum of those coefficients and divide them by how many there are, which is 10. You can see what @Amanda Kalinowska did in the comments section to see it in action.
(0.35+0.42+0.25+0.21+0.31+0.38+0.36+0.41+0.46+0.31)= 3.46 (sum -> NO 1.0's!)
3.46/10 (mean)= 0.346
How did he calculate the correlation between item 1 and item 2 is 0.35?
@@kia2663 I think you could use any statistical software to compute those correlations. In R, two options are Hmisc:: rcorr(as.matrix(df)) and GGally:: ggpairs(df)
2:30 Aaaaaaand you've lost 95% of the audiences who were indeed looking for a "simple" explanation of this issue. Even assuming they are aware how to calculate corelation, how on earth do you expect them to know where you pulled all these figures out of?
Thank you so so much for these videos. I wish there was a button for I **really** like this. Incredibly clear!!
This good staff..Iam study a PHD program and this is good for my thesis
How did you know the correlation between item 1 and item 2 is .35?
Does " the average correlation between the items mean the average of the items"?
Very concise and clear instructional video, Thanks
How did you get the .346?
he added up all of the correlations (that aren't just something correlated with itself) then divided by the number of correlations.
Hello, can you please explain how you got .346? Which numbers did you use to find the mean. I got a mean of .446 instead of .336. Please help! :) Thank you!
+Jannette Reyes I think its the average of all correlation , except for correlation of the same item (equal to 1). (all no. in the table) hope it helped =)
how about the 2.384?
??
I don't know whether this is the right way to calculate or not, but I got .346 by this way... (.35+.42+.25+.21+.31+.38+.36+.41+.46+.31)/10= 3.46/10= .346.
@@nirobkothopokothon Ah, yes! You don't include the 1's because that's the correlation value between a variable and itself. Most statistical packages show the 1's but they're not correlations between 2 different variables.
Can you explain how you got .346?
Awesome explanation … thank you so much !
Awesome class … thank you so much .
Thank you so much for this video, it is very helpful
Can I take the same procedure when I use binominal scores (ex:0 or 1)?
How do you get the .346 number?
it is amazing! maaaaany thanks for your explanations!
Hello. If the items are ordinal (e.g., a Likert scale) do you use Spearman correlation?
+Joseph Wehbe No. In such a case you should use polychoric correlations, which would give you ordinal Cronbach's alpha. SPSS doesn't do this analysis, unfortunately, but it can be done. Check out:
Gadermann, A. M., Guhn, M., & Zumbo, B. D. (2012). Estimating ordinal reliability for Likert-type and ordinal item response data: A conceptual, empirical, and practical guide. Practical Assessment, Research & Evaluation,17(3), 1-13.
In practice, people just run Pearson correlations with Likert data for Cronbach's alpha. Doing so underestimates reliability, but it's better than Spearman correlations, in this context.
How did you get these numbers? .35 .42 .25 .21 .31 .38 .36 .41 .46 .31 Thankz
They are just random values that were input into the table to provide an example that would help demonstrate the Standard Cronbach Alpha's formula.
How do you compute for the inter item correlation?
Check out the Cronbach's alpha SPSS video for that.
how2stats ok thanks!
How does he get 2.384? I get the same as the other, 1.73. 1+(5-1)x0.346 is also 1.73 isn't it? So confused
1+ (4*.346)= 2.384
Spearman correlation???
Is cronbach alpha sufficient to say that a questionnaire is valid?
And how do u test for validity? Thanks guys.
Cronbach's alpha only provides an indication of test score reliability, not validity.
@@novachrono2236 - no because even though the items nay be consistent, they could be consistently wrong as my professor said. You have to make sure they are measuring what you intended to measure.
Many thanks dear!
thanks for sharing information!
Can you explain about r bar
r bar = mean inter-item correlation
I hate statistics, I really hate everything related to the numbers.
hello can you explain how did you get 0.346? coz i get 3.46 when i compute and i dont know to make it 0.346.
divided by N which in this case is 10
What @Richii said. To make it clearer, there are 10 coefficients (all the floating point values, which are the values with decimal points, except for the '1.0' values). So, take the sum of those coefficients and divide them by how many there are, which is 10. You can see what @Amanda Kalinowska did in the comments section to see it in action.
thank you!
where did he get .346?
It's the average correlation.
Going so fucking slow!!