Calculate p-value for one-sample T-test in SPSS: 1 - p/2? In SPPS, for the p-value in a one-sample T-test, 1-tailed:
When you find a mean that is in the opposite direction of your alternative hypothesis, do you have to do 1-p/2 or just p/2?
Thank you!
 A: Results from R: Consider a random sample of size $n=40$ from
the distribution $\mathsf{Norm}(\mu=100, \sigma=15).$
Begin with numerical descriptive statistics:
set.seed(2021)
x = rnorm(40, 100, 15)
summary(x);  sd(x);  length(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  70.54   82.79   99.45   98.58  113.54  125.95 
[1] 17.13621  # sample standard deviation
[1] 40        # sample size


You can compute one-sample t tests using $n = 40, \bar X = 98.58,$ and
$S_x= 17.126.$ I will use R to get P-values for a two-sided test, and
two one-sided tests.
Two-sided test: $H_0: \mu = 105$ against $H_a: \mu \ne 105.$
The hypothetical mean is $\mu = 105$ and the sample mean is $\bar X = 98.58.$ Is the sample mean significantly different from the hypothetical mean?
t.test(x, mu=105)$p.val
[1] 0.02286777

The P-value is $0.0229,$ rounded to four places. The sample mean is
significantly different from the population mean at the 5% level of significance because $P = 0.0229 < 0.05 = 5\%.$ The null hypothesis is rejected.
One-sided test: $H_0: \mu = 105$ against $H_a: \mu < 105.$
The hypothetical mean is $\mu = 105$ and the sample mean is $\bar X = 98.58.$ Is the sample mean significantly smaller than the hypothetical mean?
t.test(x, mu=105, alt="less")$p.val
[1] 0.01143389

The P-value is $0.0114,$ rounded to four places. [Within rounding error,
this is half of the P-value of the two-sided test.] The sample mean is
significantly smaller than the population mean at the 5% level of significance because $P = 0.0114 < 0.05 = 5\%.$  The null hypothesis is rejected.
One-sided test ("opposite direction"):  $H_0: \mu = 105$ against $H_a: \mu > 105.$
The hypothetical mean is $\mu = 105$ and the sample mean is $\bar X = 98.58.$ There is no evidence that the sample mean significantly larger than the hypothetical mean. So we are not going to be able to reject $H_0.$
t.test(x, mu=105, alt="greater")$p.val
[1] 0.9885661

The P-value is $0.9986$ rounded to four places. [This is one minus the P-value of the useful one-sided test above.] The sample mean is not
significantly larger than the population mean at the 5% level of significance because $P = 0.9986 > 0.05 = 5\%.$ (As was obvious from above,
$H_0$ is not rejected.)
