Calculate the 95% confidence intervals for the slope and y-intercept determined in Example 5.10.

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Accepted Answer

Again, as you work through this example, remember that x represents the concentration of analyte in the standards [latex](C_S)[/latex], and y corresponds to the signal [latex](S_{meas})[/latex]. To begin with, it is necessary to calculate the standard deviation about the regression. This requires that we first calculate the predicted signals, [latex]\hat y_i[/latex], using the slope and y-intercept determined in Example 5.10. Taking the first standard as an example, the predicted signal is

[latex]\hat y_i= b_0 + b_1x = 0.209 + (120.706)(0.100) = 12.280[/latex]

The results for all six solutions are shown in the following table.

[latex]x_i[/latex]	[latex]y_i[/latex]	[latex]\hat y_i[/latex]	[latex](y_i-\hat y_i)^2[/latex]
0.000	0.00	0.209	0.0437
0.100	12.36	12.280	0.0064
0.200	24.83	24.350	0.2304
0.300	35.91	36.421	0.2611
0.400	48.79	48.491	0.0894
0.500	60.42	60.562	0.0202

Adding together the data in the last column gives the numerator of equation 5.15, [latex]\sum (y_i-\hat y_i)^2[/latex] , as 0.6512. The standard deviation about the regression, therefore, is

[latex]s_r=\frac{\sum\limits_{i=1}^{n}(y_i-\hat{y_i} )^2}{n-2}[/latex] 5.15

[latex]s_r = \sqrt{\frac{0.6512}{6-2} } =0.4035[/latex]

Next we calculate [latex]s_{b_1} and s_{b_0}[/latex] using equations 5.16 and 5.17. Values for the summation terms [latex]Σx_i² and Σx_i[/latex] are found in Example 5.10.

[latex]s_{b_1}=\sqrt{\frac{ns_r^2}{n\sum{x_i^2}-(\sum{x_i} )^2 } } =\sqrt{\frac{s_r^2}{\sum({x_i}-{\bar{x} } )^2 } }[/latex] (5.16)

[latex]s_{b_0}=\sqrt{\frac{s_r^2\sum{x_i^2} }{n\sum{x_i^2}-(\sum{x_i} )^2 } } =\sqrt{\frac{s_r^2\sum{x_i^2} }{n\sum{(x_i}-\bar{x} )^2 } }[/latex] (5.17)

[latex]s_{b_1}=\sqrt{\frac{ns_r^2}{n\sum{x_i^2}-(\sum{x_i} )^2 } } =\sqrt{\frac{(6)(0.4035)^2}{(6)(0.550)-(1.500)^2} } =0.965[/latex]

[latex]s_{b_0}=\sqrt{\frac{s_r^2\sum{x_i^2} }{n\sum{x_i^2}-(\sum{x_i} )^2 } } =\sqrt{\frac{(0.4035)^2(0.550)}{(6)(0.550)-(1.500)^2} } =0.292[/latex]

Finally, the 95% confidence intervals (α = 0.05, 4 degrees of freedom) for the slope and y-intercept are

[latex]β_1 = b_1 ± ts_{b_1} = 120.706 ± (2.78)(0.965) = 120.7 ± 2.7[/latex]

[latex]β_0 = b_0 ± ts_{b_0} = 0.209 ± (2.78)(0.292) = 0.2 ± 0.8[/latex]

The standard deviation about the regression, [latex]s_r[/latex], suggests that the measured signals are precise to only the first decimal place. For this reason, we report the slope and intercept to only a single decimal place.

Question 5.11: Calculate the 95% confidence intervals for the slope and y-i...

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