My answer will address your questions more specifically, though I have no objections to the response articulated by @AJKOER. You should also investigate these effects in a regression framework. It is more flexible and you can perform a sub-group analysis by year.
It looks like we can almost surely say that this higher mean of the sample of awarded movies is not accidental and that movies with award winning cast score higher.
Yes. The output suggests your results are not due to chance. There is strong evidence that the population mean difference is different from 0. You do not appear partial to any one side, so you care about a difference in either direction.
But then, for example, the golden globe sample consists only of 76 rows, wouldn't that make this test unreliable?
No. The group sample sizes are sufficient. In general, there is no minimum sample size for a $t$-test to be valid. This post should address most of your concerns related to sample size.
Also, I see that the confidence interval is negative, which seems a bit weird.
Your confusion concerns the ordering of your variables inside the t.test()
function. In your example, the average rating of the award-winning movies is larger than the non-award winning movies. Based upon your ordering scheme, your difference in sample means is negative, and your confidence interval is bounded around that estimate. In fact, the difference in means is significantly bounded away from zero—to the left. If you reverse the order, your difference in sample means would be positive, which would consequently shift your interval to the right of zero. To demonstrate this explicitly:
$$
\overbrace{6.301449}^{\text{Non-Award}} - \overbrace{7.422947}^{\text{Award}} = \underbrace{-1.121498}_{[-1.236719\hspace{1mm},\hspace{1mm} -1.006276]}
$$
In the foregoing example, the 95% confidence interval is entirely to the left of zero. However, if you reverse the order of your samples inside of the function, then your confidence interval will reflect about zero. Reversing the order results in the following:
$$
\overbrace{7.422947}^{\text{Award}} - \overbrace{6.301449}^{\text{Non-Award}} = \underbrace{1.121498}_{[1.006276\hspace{1mm},\hspace{1mm}1.236719]}
$$
Variable ordering determines which side of 0 your interval will lie. Since the interval is bounded away from zero to the left, it is also bounded away from zero to the right. You can tell the same story with either approach. You can test this below using a basic simulation in R. See the code at the bottom of this answer.
Does it mean that most years movies had an average score of 5.3 or lower?
Not quite. This is not a test of "most year's movies" in your sample. Rather, you are estimating a difference in two means (i.e., $\bar{X}_{1} - \bar{X}_{2}$). Based upon your output, you can be 95% confident that the true mean difference in movie ratings lies somewhere between -1.236719 and -1.006276 units. I use the term units, but you should tailor it the appropriate rating scheme.
Also, would it be possible to run randomly generated sample that consists of 76 numbers from 0 to 10 and replicate it 10,000 times and then say whether the globe's mean is just accidental or reliable?
R's replicate()
function should work just fine. You could do the following:
replicate(10000, t.test(..., ...))
,
where each ...
is a different sample. This online resource may be helpful!
# Ordering the Sample Means
set.seed(13)
x = rnorm(10)
y = rnorm(10) + 5
t1 <- t.test(x, y) # x - y
t2 <- t.test(y, x) # y - x
t1$conf.int # To the "left" of zero
t2$conf.int # To the "right" of zero
t.test
. $\endgroup$