Paired Observation Hypothesis Testing I'm doing a project presentation for applied statistics about hypothesis testing. Me and my partner claimed that the average amount of money that people spend per month on groceries is less than or equal to the average amount spent on dining out. We've used survey monkey (a free online survey tool) to receive data from people and ended up receiving 34 responses. 
The survey asks the average amount of money per month that they spend on groceries and dining out. This is a paired observation therefore I used a paired t-test to draw conclusion from the data. 
I have two questions:
1.) The formula has d0, and d0 = u1 - u2. Now I'm not sure if the value that I gave 
d0 = 0 is correct. d0 would definitely be equal to zero if my null hypothesis was u1 = u2, since the difference between the mews will be zero. 
2.) Is it okay to assume that the population of the amount of money spent per month on groceries and dining out are normally distributed? Even though I don't know the real shape of the two population distributions? 
Any thoughts or comments about this? The image below contains my work and conclusion. The image contains my work and the conclusion.
Also below is the data that we got from the survey. The green is the difference between the average money spent on groceries and dining out. 

Here is my updated work and conclusion 

 A: Your experiment does collect data in a paired sense, so a paired test of the difference is appropriate for testing your hypothesis.
The assumption of normality could be assumed as you have done, but even if this assumption is mildly violated in reality, the paired $t$-test should be reasonably robust for a sample size of $n = 34$ pairs.  Alternatively, you could apply a nonparametric test, such as the Wilcoxon signed rank test.
The more serious issue with your work, however, is a misunderstanding of the nature of hypothesis testing.  Think of a hypothesis test like a criminal trial:  a defendant accused of a crime is presumed innocent, and the prosecution's goal is to furnish sufficient evidence that a reasonable person would conclude that this initial presumption of innocence is incorrect--that is, the defendant is guilty beyond a reasonable doubt.
With this analogy in mind, the function of the null hypothesis $H_0$ is to serve as the "presumption of innocence."  The alternative hypothesis, $H_a$, is the conclusion of guilt with a high degree of confidence.  And in a criminal case, a failure to prove guilt does not always mean the defendant is innocent--it simply means the standard of proof was not met.  Thus, a failure to reject $H_0$ in favor of $H_a$ does not mean the null hypothesis is true:  it simply means the evidence was not strong enough to conclude $H_a$.
In a hypothesis test, we never say "accept $H_0$."  The decision is always between one of these two choices:


*

*Reject $H_0$ in favor of $H_a$ (with $100(1-\alpha)\%$ confidence, i.e., a significance level of $\alpha$).

*Fail to reject $H_0$:  the test is inconclusive.
Now, if your original hypothesis is a belief that people spend less on groceries than on dining out, then the burden of proof is on this claim, consequently, this must be your alternative hypothesis, not the null.  So your hypothesis must have the structure $$H_0 : \mu_g \ge \mu_d, \quad \text{vs.} \quad H_a : \mu_g < \mu_d,$$  where $\mu_g$ is the mean amount spent per month on groceries, and $\mu_d$ the mean amount spent per month on dining out.  If the test statistic you calculate (either with or without assuming normality) leads you to reject $H_0$, then, and only then, can you affirm that people spend less on groceries than on dining out.  Otherwise, your test is inconclusive.
Why is there this apparent asymmetry between $H_0$ and $H_a$?  The reason is because the calculation of the test statistic occurs under the assumption that $H_0$ is true.  Consequently, the $p$-value of the test is a conditional probability of obtaining a sample at least as extreme as the one you observed, given that the null hypothesis is true.  And this is why you can never "accept $H_0$:"  it makes no sense to say "the null hypothesis is true, assuming it is true."  That's circular reasoning.
Finally, I advise that you recompute the paired differences in your list, and double check the accuracy of the data you collected.  You should be clear that the paired difference is calculated within each respondent, i.e., if a respondent said they spend $x_i = 500$/month on groceries and $y_i = 750$/month dining out, their difference would be $d_i = x_i - y_i = -250$, and your test statistic would need to be negative in order to reject $H_0$.
A: Unfortunately, your data are only barely visible on my screen.
I have tried twice to get the same $\bar d$ and $s_d$ you did,
but without success. (Close, but not the same.) Here are the differences as I read them:
 d = c(120,-50,200,250,-100,0,200,650,800,480,
       30,40,20,110,-130,320,50,-50,-300,100,
       200,-100,100,55,250,-5,250,200,300,85,
       250,720,250,200)

A boxplot of these data is shown below. As is typical of dollar amounts
for expenditures, there is some skewness toward the right with a
few (mild) outliers on the right, as shown in the boxplot. Because of the skewness sample (and population) means are larger than corresponding medians.

The 'notches' on the sides of the boxplot indicate a rough nonparametric
confidence interval for the median of the population from
which the data were sampled. ('Nonparametric' means that no
assumption about normality of the data was made.) Because this
interval $(53, 177)$ does not include $0$ it seems plausible
that people spend more on groceries than in restaurants.
A 95% CI for the population mean difference is of the form
$\bar d \pm 2.03 s_d / \sqrt{n},$ where 2.03 cuts 2.5% of the
probability from the upper tail of Student's t distribution
with $df = 33$. Using my values of $\bar d$ and $s_d$, the CI is about
$(80, 243).$ Because the data are not quite normal, the endpoints
of this interval are only approximate. But not so far off that
the CI could include $0$. So it is pretty clear that people in
the population from which you sampled spend more on groceries.
(Using your $\bar d$ and $s_d$, you
will get a slightly different CI, but not enough to change
the interpretation.)
A nonparametric Wilcoxon 95% CI for the population median is
approximately $(74, 225),$ using my version of the data. Again,
clearly more money spent on groceries.
(It is not possible to get an exact result because there are several
sets of
tied values among the differences.) 
The people who respond to Survey Monkey polls may not have
the same eating habits as the population in general, but this
is a nice project. Much more interesting than working with
an exercise in a textbook where the data are made up and
have little or nothing to do with anybody's reality.
