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I have some confusion in interpreting p-value from SPSS.

If my hypothesis is one-tailed, after getting the p-value from SPSS, since the p-value from SPSS is always two-tailed,

ie. p = 0.06 (from SPSS), alpha = 0.05

In this case, does it means that my p-value will become 0.03 and significant? or does it means that I need to use p < 0.025 as my cut-off point to consider the result as significant?

Thanks a lot.

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    $\begingroup$ You may find this link helpful: stats.idre.ucla.edu/other/mult-pkg/faq/general/…. The link explains how to derive a one-sided p-value from a two-sided p-value based on STATA output. The derivation requires that you know the direction of the alternative hypothesis and assumes that the sampling distribution of the test statistic under the null hypothesis is symmetric about zero. $\endgroup$ Commented Jan 5, 2019 at 20:02
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    $\begingroup$ yup, i have actually read this before asking but still confused, because in real life it is very hard to get sampling distribution that is perfectly normally distributed, so was thinking in that case, if just divide the p-value by 2 it will be not reasonable as the distribution won't be symmetric at both sides, now i think i kinda get it, thanks a lot for the help $\endgroup$
    – Cheah
    Commented Jan 6, 2019 at 7:14
  • $\begingroup$ For "large-ish" n as in my Answer, the distribution of the t statistic tends to have "reasonably" close to a t distribution; maybe not exactly symmetrical, but the P-value of a directional t test (in the correct direction) is about half that of a two-sided test. – BruceET 16 mins ago $\endgroup$
    – BruceET
    Commented Jan 6, 2019 at 17:44

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Generally speaking, if the p value is calculated from a two tailed test you will need to divide it by two if you want to use it for a one tailed test as the area under the probability curve is divided by two as well. enter image description here

In the picture you can see the area "used" for a right tailed test. A p value in a two sided test is the marginal significance of both sides added together. So in a two tailed test with a p-value of 0.02 the marginal significance on the right and left side would both be 1%.

With that being said, I would be careful with using the p-value of a two tailed test to reject the null of a one tailed test. If you simply say 0.03 < 0.05 thus you can reject the H0 then you could reject the H0 for both, a right sided and a left sided test. In terms of interpretation this does not make sense to me as you would say the estimate is significantly larger than the H0 while at the same time being significantly smaller than the H0.

Instead I would recommend using the t-values for your test decision. More specifically, if the t-value is larger than then critical t-value (the t value obtained at a 95% area under the probability curve (or the t-value of two sided test with alpha = 0.1) and the degrees of freedom of your test) and you do a right sided test then you can reject the H0. Note that you do not take the absolute t-value if you do a one tailed test. For a left sided test you can reject the H0 if the t-value is smaller than the critical t-value (at 5%) or, if you feel lazy, just -(the critical t-value at the 95% level from the right sided test) since the t and z distributions are symmetric.

I don't use SPSS so I don't know whether it actually gives you a two-tailed test p-value if you define the H1 and H0 as the proper side one tailed test.

Hope this helps.

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  • $\begingroup$ (+1) for nice figure, illustrating the symmetry of the null distribution. And welcome! But you don't explicitly caution that one needs to be sure the direction of the one-sided alternative is correct in order to get the P-value by dividing by 2. $\endgroup$
    – BruceET
    Commented Jan 6, 2019 at 1:31
  • $\begingroup$ Thanks for the suggestion, so far we were asked to decide the significance level based on p-value only, which is why I have the confusion, because it looks like adjusting the p-value and making it significant when two-tailed test is not significant, i guess it will be risky to make conclusion like this. Will try to explore further into t-values, thanks a lot. P.S. SPSS only gives p-value for two-tailed test. $\endgroup$
    – Cheah
    Commented Jan 6, 2019 at 7:51
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Data and description. Here is a random normal sample $X_1, X_2, \dots, X_{40}$ generated using R statistical software. We wish to test $H_0: \mu= 100$ against $H_1: \mu > 100.$

set.seed(2019)
x = rnorm(40, 105, 10)
summary(x)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   82.37   96.60  102.21  103.69  112.38  131.36 

stripchart(x, pch="|")
 abline(v=100, col="red", lty="dotted")
 abline(v=mean(x), col="blue", lwd=2)

stripchart

From the stripchart we see the locations of the $n = 40$ individual observations and their sample mean $\bar X = 103.69$ (heavy blue vertical line). The null hypothesis $\mu_0 = 100$ is shown as a dotted red line. We can see that $\bar X > \mu_0,$ but the question is whether $\bar X$ is significantly greater than $\mu_0.$

Two-sided test. In R a two-sided test $H_0: \mu =100$ against $H_1: \mu \ne 100.$

t.test(x, mu=100)

        One Sample t-test

data:  x
t = 2.1772, df = 39, p-value = 0.03558
alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
 100.2619 107.1209
sample estimates:
mean of x 
 103.6914 

So the two-sided test shows that $\bar X$ is significantly different from $\mu_0 = 100$ with P-value 0.036, so we can reject at level $\alpha = 4\%,$ but not at level $\alpha = 3\%.$

Sensible right-sided test. For our right-sided test of $H_0: \mu= 100$ against $H_1: \mu > 100,$ the P-value is half as large. That is, the P-value is $0.03558/2 = 0.01779.$ Thus, $\bar X$ is significantly greater than $\mu_0 = 100$ at level $\alpha = 2\%,$ but not at level $\alpha = 1.5\%.$

In R statistical software there can be no confusion about this, because one can explicitly do a one-sided test.

t.test(x, mu=100, alt="greater")  # Note the parameter 'alt="greater"'

        One Sample t-test

data:  x
t = 2.1772, df = 39, p-value = 0.01779
alternative hypothesis: true mean is greater than 100
95 percent confidence interval:
 100.8347      Inf                # Note one sided CI with lower bound
sample estimates:
mean of x 
 103.6914 

Absurd left-sided test. However, in this situation it would make no sense to do the one-sided test $H_0: \mu= 100$ against $H_1: \mu < 100,$ because with $\bar X = 103.69,$ we have evidence that $\mu > 100$ if it is different from 100. In particular, you cannot divide the two-sided P-value in half to get a P-value for this nonsense alternative test.

In R, if you insist on trying to do the left-sided test, here is what you get:

t.test(x, mu=100, alt="less")

        One Sample t-test

data:  x
t = 2.1772, df = 39, p-value = 0.9822    # Note absurdly high P-value
alternative hypothesis: true mean is less than 100
95 percent confidence interval:
     -Inf 106.5482                       # Note CI with bound above x-bar 
sample estimates:
mean of x 
 103.6914 

In hypothesis testing, one should always be wary of a P-value that is near $1.$ Almost always, this is an indication that one is trying to do something stupid. "If the P-value is near 0, doubt the null hypothesis; if the P-value is near 1, doubt the model." In @IsabellaGhement's fine Comment, the point is made that you need to know the direction of the alternative hypothesis.

Notes: (1) In this Answer data were generated from a normal population with mean $\mu = 105,$ so there is no doubt that $\mu > 100.$ Of course, in a real application you would not have such information. (2) It puzzles me that SPSS designers have chosen not to give P-values for one-sided t tests. You are not the first person ever to be confused by this.

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  • $\begingroup$ Have much better understanding on this, thanks a lot for the clear explanation! $\endgroup$
    – Cheah
    Commented Jan 6, 2019 at 7:58

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