How to use R to determine if a value is statistically LARGER than an average? So I have daily temperature data for each day of the year going back up to 100 years for different locations. With this, I have created a mean value for each day of the year and a 95th percentile again for each day of the year. 
I am comparing these values to the daily readings for a specific year (e.g.2016). However, I am only interested if the 2016 values are greater than the daily mean/95th percentile. 
Is there a way to determine if the values are significantly greater than the mean/95th percentile?
In essence, there are 366 values for each column (one for each day of the year), and the mean/95th percentile and the 2016 data would be paired.
A rough idea of what my data looks like:
DAY OF YEAR: 1, 2, 3 
MEAN: 12, 10, 11
2016 Temp: 17, 20.6, 13
Then:
DAY OF YEAR: 1, 2, 3 
95th Percentile: 16, 14, 13
2016 Temp: 17, 20.6, 13
Apologies if the formatting is confusing.
So, all the 2016 values are larger than they 'should' be, but is this significant?
Is there a way to determine if the values are significantly greater than the mean/95th percentile, not just statistically different?
 A: Suppose you have $n = 20$ observations for the maximum temperature on
a particular day of the year during a period when you believe the
mean of such temperatures was stable at $\mu.$  Perhaps the sample mean and SD
for the twenty-year period are $\bar X = 35, S = 10.$
Then a 95% CI for the true population $\mu$ is $\bar X \pm t^*S/\sqrt{n},$
where $t^*$ cuts probability $0.025$ from the upper tail of Student's t
distribution with $\nu = n-1 = 19$ degrees of freedom. Specifically,
for the data above a 95% CI for $\mu$ is $35 \pm 2.09(10/\sqrt{20}),$
which is $35 \pm 6.619$ of $(28.381, 41.619).$ [Computations in R.]
q = qt(.975, 19); q
[1] 2.093024
me = q*10/sqrt(10); me
[1] 6.618723
pm = c(-1,1);  35 + pm*me
[1] 28.38128 41.61872

Now you have a new high temperature 45 for that day this year.
Compared with the historical distribution, estimated to be
$\mathsf{Norm}(35,10),$ this new temperature would be at about the 84th percentile. That's certainly higher than 'average', but not really
very exceptional.
pnorm(45, 35, 10)
[1] 0.8413447

It would not be fair to compare the new one-time high temperature 45 with the
CI $(28.381, 41.619)$ for the historical $\mu.$ (Your new temperature of 45 is an observation, not a parameter.)
However, based on the 20 years of data, we can get a prediction interval
for an additional observation. It is of the form $\bar X \pm 2.09S\sqrt{1 + \frac{1}{20}},$ which is $34 \pm 21.45$ or $(13.55, 56.45).$ So the new high temperature is within the interval of values for an additional observation
from the historical distribution. [If you get a year with a high temperature of 60 for that day, then you have an unusual observation according to the 95% prediction interval.]
ME.pred = q*10*sqrt(1+1/20);  ME.pred
[1] 21.44711
35 + pm*ME.pred
[1] 13.55289 56.44711

However, if you get five years of temperatures for that day that average well
above $35,$ you could do a Welch 2-sample t test to compare the 20-year
historical data with the new data, to see if the new mean is significantly
higher than the historical one.
A: Welcome to Cross Validated! 
To answer your question: how do you define "significantly greater"? It appears one easy thing to do would be to calculate z value given the 2016 observation and define a z-threshold above which you call all observations as "significantly different". There are a number of ways to define and detect such "outliers" and you could look into those as well.
A: Suppose that $T_{dy}$ is the temperature in a particular (day,year). Let $\alpha_d$ denote a day-specific fixed effect, $\mathbb{1}\{ y = 2016 \}$ an indicator for 2016, and let $\varepsilon_{dy}$ be an error term.
$$ T_{dy} = \beta_0 + \beta_1 \mathbb{1}\{ y = 2016 \} + \alpha_d + \varepsilon_{dy} $$
We can test whether the mean temperatures in 2016 are larger than average (after controlling for day-specific fixed effects, by testing whether $\beta_1 \ne 0$.
More generally what you are doing is "testing" a structural break in 2016 with respect to previous years.
You can test this in R by running the following code
# Import the "plm" package
library('plm')

# Your data: temperaturedata
# Has three columns: "day", "year" and "temperature".

# Create a dummy variable for the year
data$y2016= ( data$year == 2016 )

temperaturemodel <- plm(temperature ~ y2016, 
                      data = temperaturedata,
                      index = c("day"), 
                      model = "within")

# print summary using autocorrelation-robust standard errors
coeftest(temperaturemodel, vcov. =vcovNW)

The option "vcovNW" computes Newey and West (1987) autocorrelation-robust standard errors.


