Which t-test should I go for? Newbie here. I'm wondering which t-test I should go for: a one-sample t-test or a paired-sample t-test? Happy to receive other suggestions.
A bit about my data:
I have data from 18 participants on how much time they planned to work on a Monday, and how much time they actually ended up working (from two different time-use diaries).
I checked and the data is normally distributed.
What I want to show with the data is that there is a difference between the amount of time people planned to work, and the amount of work they were eventually able to do (except for Diary Participant 11 who worked for exactly the amount of time they had planned).

My question is: which test would you say is more appropriate for this data, if any?
 A: It can be a paired t test using columns 'Planned' and 'Actual', or (equivalently) a one-sample t test using column 'Difference'.
I wrote your differences into the vector d below. (Maybe check my
typing.)
d = c(140, -660, -60, -110,  195, 455, -140, -110,
      -210, -30, 0, 40,  -60, 230, 65, -55,
      -90, -235, - 35)

The first question is whether these differences are normally distributed.
A normal probability plot is "mainly" linear with a few exceptions, especially the lowest (-660) and highest (455) values .
qqnorm(d);  qqline(d, col="green2")


These extreme values show as 'outliers' in a boxplot of the differences.
boxplot(d, horizontal=T, pch="|")


Nevertheless, a Shapiro-Wilk test for normality does not (quite) reject the
null hypothesis (normal data) at the 5% level.
shapiro.test(d)

         Shapiro-Wilk normality test

data:  d
W = 0.91394, p-value = 0.08748

Various statisticians have different favorite ways of judging normality and tolerance for departure from normality in a one-sample t test.
A one-sample t test on the differences, does not find the average difference
$\bar D = -35.26$ in the sample mean to be significantly different from $0$
at the 5% level of significance. [If you had done a paired t test in R, the output would have looked much the same; the first step of the paired procedure would have been to find the differences used in the output below.]
t.test(d)

        One Sample t-test

data:  d
t = -0.68782, df = 18, p-value = 0.5003
 alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
  -142.97372   72.44741
sample estimates:
mean of x 
-35.26316 

If someone disbelieves the accuracy of the t test (on account of non-normality), they might say a nonparametric paired
Wilcoxon test (equivalently, one-sample Wilcoxon signed-rank test)
would be better. An objection to using this Wilcoxon test is that the
data are not symmetrical. (Look at the boxplot.) However, the Wilcoxon SR test also fails to reject the null hypothesis that subjects deviated
significantly from plan. (P-value above $0.05 = 5\%.$%
wilcox.test(d)

        Wilcoxon signed rank test with continuity correction

data:  d
V = 66.5, p-value = 0.4203
alternative hypothesis: true location is not equal to 0

As often happens in practice, neither test is perfect. But arguments
can be made for using either. And in the end, neither test finds a
significant difference from planned work time.
Pick one of the two tests
an use it. Neither test shows results anywhere near the borderline for rejection.
Some of your workers aren't very good at sticking to plan, but they do
not show a strong trend in either direction--above or below plan.
Note: For completeness: one possible additional test comes to mind. It's called a sign test.
Out of $n = 19$ observations, you have $n^\prime = 18$ non-zero results. Of these non-zero tests 6 are positive and 12 are negative. The probability of 6 or fewer positives or of 12 or more negatives
out of 18 is the P-value $0.2379$ of the sign test--nowhere near significant at the 5% level. There are also objections to the sign test because data are not symmetrical and because looking just at signs 'throws away' too much information.)
sum(dbinom(c(0:6,12:18), 18, .5))
[1] 0.2378845

By way of confirmation, Minitab statistical software shows the following results for
a sign test on your data:
Sign Test for Median: d 

 Sign test of median =  0.00000 versus ≠ 0.00000

    N  Below  Equal  Above       P  Median
d  19     12      1      6  0.2379  -60.00

