Methods used to compare two spectra in order to determine the effects of irradiation of sample I would like to compare two absorption spectra (or interferograms) and conclude whether between these two there are statistically significant differences at particular wavelength intervals. At the moment, I have data of two experiments that look like this:
    # A tibble: 6 x 5
      t     x1     y1     x2     y2
  &lt;dbl&gt;  &lt;dbl&gt;  &lt;dbl&gt;  &lt;dbl&gt;  &lt;dbl&gt;
1 3999. 0.0124 0.0132 0.0122 0.0113
2 3998. 0.0125 0.0130 0.0122 0.0116
3 3997. 0.0122 0.0131 0.0122 0.0113
4 3996. 0.0121 0.0136 0.0122 0.0114
5 3995. 0.0124 0.0139 0.0122 0.0122
6 3994. 0.0125 0.0141 0.0122 0.0129

The first column represents the wavenumber, the x columns represent the absorbance of sample and the y columns represent the absorbance of irradiated sample (before and after).
I was wondering whether I could compare these data (x and y) as time series and if so, what could be the method to quantify the differences, if any, between the samples before and after irradiation. Maybe it's already been done and there is somewhere some information as to how to compare the spectra if the wavenumber is interpreted as time ($x$ axis).
I did the t-test in R and in both experiments the null hypothesis could not be rejected, although for the second experiment (x2, y2) the $p$ value was much lower than for the first. If I average the x and y, and then plot both data, I see that there are visible differences at certain wavelength intervals. But how can I check for sure the differences between spectrums?
Here is a project with similar experiments by Zezell et al. [1]. For statistical analysis they use ANOVA and Tukey's test, but how do I do it for the vectored data?
Reference

Zezell, D. M.; Benetti, C.; Veloso, M. N.; Castro, P. A. A.; Ana, P. A. FTIR Spectroscopy Revealing the Effects of Laser and Ionizing Radiation on Biological Hard Tissues. J. Braz. Chem. Soc 2015. DOI: 10.5935/0103-5053.20150246.
The table below illustrates a possible screen, values computed displayed are rounded to four decimals.  The suggest is to use the observations prior the intended irradiation ($x_1, x_2$) separate from the observations after the intended irradiation ($y_1, y_2$).  Per wavelength

compute for $x$ and $y$ the arithmetical mean value
determine the standard deviation of this sample (e.g., $x_1$ and $x_2$), the one pocket calculators sometimes label with $\sigma(n-1)$.  The table doesn't include it, but you may compute the confidence interval with a reasonable $t$ value.  Because there were only two realizations, the degree of freedoms ($f = n - 1$) equates to 1, and an understandably high $t_{1, 0.95} = 12.71$ to map the interval of

$$\bar{y} - \frac{t \cdot \sigma}{\sqrt{n}} \le \bar{y} \le \bar{y} + \frac{t \cdot \sigma}{\sqrt{n}}$$

subtract the arithmetical mean values from each other to determine the effect of the irradiation.  Expect positive as well as negative values to occur.
compute the standard deviation for this effect.  Because the effect is computed as a difference of $(\bar{x} - \bar{y})$, the error propagation of is a sum of the corresponding standard deviations. Or add the corresponding halfes of the confidence intervals about $x$ and $y$ instead.

I have no hands-on experience with R.

| lambda |     x1 |     y1 |     x2 |     y2 | mean_x | stdev_x | mean_y | stdev_y | mean_x - mean_y | stdev_x + stdev_y |
|--------+--------+--------+--------+--------+--------+---------+--------+---------+-----------------+-------------------|
|   3999 | 0.0124 | 0.0132 | 0.0122 | 0.0113 | 0.0123 |  0.0001 | 0.0123 |  0.0013 |          0.0000 |            0.0015 |
|   3998 | 0.0125 | 0.0130 | 0.0122 | 0.0116 | 0.0123 |  0.0002 | 0.0123 |  0.0010 |          0.0001 |            0.0012 |
|   3997 | 0.0122 | 0.0131 | 0.0122 | 0.0113 | 0.0122 |  0.0000 | 0.0122 |  0.0013 |          0.0000 |            0.0013 |
|   3996 | 0.0121 | 0.0136 | 0.0122 | 0.0114 | 0.0122 |  0.0001 | 0.0125 |  0.0016 |         -0.0003 |            0.0016 |
|   3995 | 0.0124 | 0.0139 | 0.0122 | 0.0122 | 0.0123 |  0.0001 | 0.0130 |  0.0012 |         -0.0007 |            0.0013 |
|   3994 | 0.0125 | 0.0141 | 0.0122 | 0.0129 | 0.0123 |  0.0002 | 0.0135 |  0.0008 |         -0.0011 |            0.0011 |