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t.test(price1,price2,mu=0,alt="two.sided",paired=T,conf.level=0.95)

    Paired t-test

data:  price1 and price2
t = 1.4268, df = 29, p-value = 0.1643
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -6.861169 38.518502

If initially I hypothesized that: $H0: \mu_1 = \mu_2$ and $Ha: \mu_1 > \mu_2$

And those were the results from my paired t-test in R, I'm not sure how to make sense of them. Because the resulting p-value is 0.1643 at significance level of $\alpha$ = 0.05, this means we accept H0 correct? I don't know how to explain the 95% confidence interval from these results either, what does the -6.861169 mean? If someone could just put all of this into words that would be easy to understand, that'd be great going forward, thanks!

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    $\begingroup$ If your alternative is one sided, why did you explicitly ask for a two-sided test in your code? Your actions don't fit with your stated alternative. $\endgroup$
    – Glen_b
    Commented May 1, 2016 at 0:54

2 Answers 2

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  1. No, we don't "accept $H_o$", instead we fail to reject the null hypothesis. So we are not "rejecting the null", implying that we don't have enough evidence to assume that the difference in means is different from zero.

  2. The confidence interval includes the value zero, and gives us the ranges of values within which the mean difference in the population would lie with a confidence level of $95\%$. In your results, this is a very broad range, but the inclusion of the value zero prevents you from excluding the null.

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  • $\begingroup$ I thought since the p-value of 0.1643 is greater than the significance level of 0.05, it would imply that we fail to reject the H0, is that not correct? $\endgroup$
    – Nick Lo
    Commented Apr 30, 2016 at 21:00
  • $\begingroup$ Exactly! We fail to reject the null hypothesis ($H_o$). The entire construct is set up from the perspective of $H_o$ and you assess how probable it would be for you to get a test statistic like the one you get under the null. If it is not that unlikely, you don't reject the null. $\endgroup$
    – Antoni Parellada
    Commented Apr 30, 2016 at 21:07
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The alternative hypothesis that you proposed is one-sided but you have used a two-sided t-test. Had you set alt="greater" in the t-test to get the one-sided result you would have had p-value = 0.082.

Do not dichotomise the result into significant and not significant on the basis of comparison of your observed p-value and the unthinkingly arbitrary threshold of 0.05. To do so is to use the "bright-line" thinking that is warned against in the recent American Statistical Association's official statement on p-values. http://amstat.tandfonline.com/doi/abs/10.1080/00031305.2016.1154108#.VyUnhmR95cw

Instead of dichotomising, look at the result and weigh it in light of what you know and what you want the data to tell you. Your data appear to contain relatively weak evidence against your null hypothesis.

There are many references in the ASA statement that will help you with the distinction between the dichotomous result of a hypothesis test and an evidence-respecting interpretation of a significance test, but I like my own: http://www.ncbi.nlm.nih.gov/pubmed/22394284

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  • $\begingroup$ The link in your profile to Melbourne University is broken, but being that you nicely identify yourself by name I took the interest in looking you up. Did I get the right person? $\endgroup$
    – Antoni Parellada
    Commented Apr 30, 2016 at 22:09
  • $\begingroup$ Yep, that's me. $\endgroup$
    – Michael Lew
    Commented May 1, 2016 at 0:09

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