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Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. We can replace the coefficients with ones that are slightly larger and smaller (according with the relative error); since z >= 0, the two resulting polynomials form estimates for the upper and lower bounds on poly[z]. (I'll use a high working precision, as in Mr.Wizard's answer, to compute the bounds precisely.)

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. We can replace the coefficients with ones that are slightly larger and smaller (according with the relative error); since z >= 0, the two resulting polynomials form estimates for the upper and lower bounds on poly[z]. (I'll use a high working precision, as in Mr.Wizard's answer, to compute the bounds precisely.)

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. We can replace the coefficients with ones that are slightly larger and smaller (according with the maximum relative error); since z >= 0, the two resulting polynomials form estimates for the upper and lower bounds on poly[z]. (I'll use a high working precision, as in Mr.Wizard's answer, to compute the bounds precisely.)

boundsPlot[error_, opts : OptionsPattern[Plot]] := With[{
    polyMin = poly[z] /. c_Real :> (1 - Sign[c] error) SetPrecision[c, Infinity],
    polyMax = poly[z] /. c_Real :> (1 + Sign[c] error) SetPrecision[c, Infinity]},
   Plot[{polyMin, polyMax}, {z, 0, 1}, WorkingPrecision -> 50, Filling -> {1 -> {2}}, opts]
   ];
bounds = boundsPlot[2^-54, PlotStyle -> {Darker@Blue, Darker@Green}, PlotRange -> 5];
absbounds = boundsPlot[2^-53, PlotRange -> 5];

plotOP = Plot[poly[z], {z, 0, 1}, PlotStyle -> {AbsoluteThickness[0.7], Red}, PlotRange -> 5];

Show[plotOP, absbounds, bounds]

A good way to think of floating-point numbers is that they represent an interval of real numbers. At best a floating-point number is an approximation obtained by rounding the true value in an interval to the binary representative. An average error of 2^-54 is about as precise as the coefficients could be if they are thought of as random variables uniformly distributed throughout each interval. The bounds above show the range of likely possible values of poly[z], although occasionally poly[z] exceeds bounds.

The upshot is this: Under the hypothesis that the coefficients are known to MachinePrecision, the OP's original plot more accurately represents what is known about poly[z], than the better-looking high-precision plot. The plot bounds seems to show the likely limits of the plot of the polynomial, under the same hypothesis (although calculating the probability is beyond me). Similarly, the plot absbounds shows absolute bounds on the poly[z]. Under the hypothesis, using absbounds or bounds seems the appropriate way to represent poly[z].

If the error is greater than 2^-53, say, because the coefficients are calculated from scientific measurements with less precision, then one can compute the corresponding bounds.

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. We can replace the coefficients with ones that are slightly larger and smaller (according with the relative error); since z >= 0, the two resulting polynomials form estimates for the upper and lower bounds on poly[z]. (I'll use a high working precision, as in Mr.Wizard's answer, to compute the bounds precisely.)

Let's show two sets of bounds: bounds is in accordance with the average relative error and will form the envelope of the OP's plot; and absbounds is in accordance with the maximum relative error.

boundsPlot[error_, opts : OptionsPattern[Plot]] := With[{
    polyMin = poly[z] /. c_Real :> (1 - Sign[c] error) SetPrecision[c, Infinity],
    polyMax = poly[z] /. c_Real :> (1 + Sign[c] error) SetPrecision[c, Infinity]},
   Plot[{polyMin, polyMax}, {z, 0, 1}, WorkingPrecision -> 50, Filling -> {1 -> {2}}, opts]
   ];
bounds = boundsPlot[2^-54, PlotStyle -> {Darker@Blue, Darker@Green}, PlotRange -> 5];
absbounds = boundsPlot[2^-53, PlotRange -> 5];

plotOP = Plot[poly[z], {z, 0, 1}, PlotStyle -> {AbsoluteThickness[0.7], Red}, PlotRange -> 5];

Show[plotOP, absbounds, bounds]

A good way to think of floating-point numbers is that they represent an interval of real numbers. At best a floating-point number is an approximation obtained by rounding the true value in an interval to the binary representative. An average error of 2^-54 is about as precise as the coefficients could be if they are thought of as random variables uniformly distributed throughout each interval. The bounds above show the range of likely possible values of poly[z], although occasionally poly[z] exceeds bounds.

The upshot is this: Under the hypothesis that the coefficients are known to MachinePrecision, the OP's original plot more accurately represents what is known about poly[z], than the better-looking high-precision plot. The plot bounds seems to show the likely limits of the plot of the polynomial, under the same hypothesis (although calculating the probability is beyond me). Similarly, the plot absbounds shows absolute bounds on the poly[z]. Under the hypothesis, using absbounds or bounds seems the appropriate way to represent poly[z].

If the error is greater than 2^-53, say, because the coefficients are calculated from scientific measurements with less precision, then one can compute the corresponding bounds.

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment. His comment is worth the emphasis, which I think the following will show. (Mr.Wizard alludes to the issue below in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. Since z >= 0, we can replace the coefficients with ones that are slightly larger and smaller (according with the maximum relative error) to compute the expected upper and lower bounds on poly[z]. (I'll use a high working precision to compute the bounds precisely as in Mr.Wizard's answer.)

The upshot is this: Under the hypothesis that the coefficients are known to MachinePrecision, the OP's original plot more accurately represents what is known about poly[z], than the better-looking high-precision plot. The plot bounds seems to show the likely limits of the plot of the polynomial, under the same hypothesis (although calculating the probability is beyond me). Similarly, the plot absbounds are certain bounds on the polynomial. Under the hypothesis, using absbounds or bounds seems the appropriate way to represent poly[z].

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.)

Let's say that machine precision is binary64, which has 53 bits of precision, and the maximum relative rounding error is 2^-53. Let's be a little fuzzy and say that the average error is roughly half that or 2^-54. The we can get almost exactly the envelope of the OP's plot as follows. We can replace the coefficients with ones that are slightly larger and smaller (according with the maximum relative error); since z >= 0, the two resulting polynomials form estimates for the upper and lower bounds on poly[z]. (I'll use a high working precision, as in Mr.Wizard's answer, to compute the bounds precisely.)

The upshot is this: Under the hypothesis that the coefficients are known to MachinePrecision, the OP's original plot more accurately represents what is known about poly[z], than the better-looking high-precision plot. The plot bounds seems to show the likely limits of the plot of the polynomial, under the same hypothesis (although calculating the probability is beyond me). Similarly, the plot absbounds shows absolute bounds on the poly[z]. Under the hypothesis, using absbounds or bounds seems the appropriate way to represent poly[z].