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  1. M

    Omitted variable bias

    I asked a teacher whether OVB and confounding are the same thing, his reply was as follows: "no, confounding and OV bias are different. OV bias occurs when you have left out a variable you should include, confounding occurs when you have multiple variables that capture the same idea in a...
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    Omitted variable bias

    Dear all, I've learned that Omitted Variable Bias happens when the missing variable is correlated with an independent variable and has causal relationship with the dependent variable. I've also learned that Confounding has the same criteria as above. My question is: is confounding a subset of...
  3. M

    Bayesian probability problem

    Oh. The formula is like this: P(Bag A) = 0.5 P(gold coin | Bag A) = 1 P(gold coin) = (150/175)
  4. M

    Bayesian probability problem

    It's calculated in this way: P(Bag A | gold coin) = (0.5 x 1) / (150/175)
  5. M

    Tossing coins

    -30 cents is expected. Who would play this game? The most he would win is only 40 cents, but he bets 50.
  6. M

    Bayesian probability problem

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  7. M

    Bayesian probability problem

    There are 2 bags. Bag A contains 100 gold coins. The other bag B contains 50 gold coins and 25 silver coins. Suppose John chose a bag and then picked 1 coin from within, the coin was a gold coin, what's the probability that he picked bag A ? The answer is 0.5833 using Bayes' Theorem. What if...
  8. M

    Performance evaluation

    Yes, you are right, the state average is known, it is something that can be derived from the data. Yes, that's my goal - to check if any compliance is significantly different from the average. For example, 67.7 % is not significantly different from 68%, while 2% is. My question is: what test...
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    Performance evaluation

    Dear all, I am doing a evaluation of performance for a group of clubs, the aim is to check which club has compliant % below the state average. The table below shows an example. Each row represents 1 club. The last row represents the state, 68% is the state average. My question is: what...
  10. M

    Statistics for bridges with wooden poles underneath

    Interesting problem. It will require a lot more than 35 sample poles to give you confidence (more than 300), but if it is average you are after, then it would require less.
  11. M

    Likert Type Items - Analysis

    Dear all, Aileen Fink said that we can treat likert type item values (which are ordinal) as if they are continuous data when we analyse them. Could we also apply this principle to descriptive statistics? For example, a question has [strongly disagree, disagree, neutral, agree, strongly...
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    Dependent t-test with extra information provided

    Oh, I see, now I can clearly see your point. I didn't know that, stats books don't mention that.
  13. M

    Dependent t-test with extra information provided

    Yes, I would conclude there's NO EVIDENCE that there is a difference between before and after measurements. Since 20 is the upper bound of normalcy. So what's the problem?
  14. M

    Dependent t-test with extra information provided

    I was trying to perform a one tailed t-test, with Ho: mu <=20 (given that mu can go up to 20 and still be considered normal). Isn't this correct?
  15. M

    Dependent t-test with extra information provided

    I recalculated it, it is still not significant under Ho: mu <=20. The t-value is negative, but the Critical Value is 1.729 (df=19). So cannot reject Ho. It was 1 tailed.
  16. M

    Dependent t-test with extra information provided

    This image illustrates my calculation. The p value I got from the test is 0.000, and the test is a two tailed test with test value=20. That p value is represented by the far right orange-coloured area. It is a p value from a two tailed test, but I am doing a one tailed test (Ho: mu < 20), so my...
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    Dependent t-test with extra information provided

    I don't think so, I am doing a one tailed test now, and 0.000 is the probability of getting >20.
  18. M

    Dependent t-test with extra information provided

    Can I calculate like this : The p value from SPSS is <0.001, so half of that will be <0.0005, that's the probability of mean >20. So the probability of mean <=20 is (1 - 0.0005).
  19. M

    Dependent t-test with extra information provided

    Oh, sorry I was wrong. His claim applied to the mean. So you were right. So I did a one tailed test, Ho is "mean difference <20". The SPSS test I used was one sample t test, I set the test value = 20. The resulting p value = 0.000 (see images below). I am wondering: how should I do the rest ...