For measurement giving y value of 2.58 , the corresponding x value and its standard uncertainty has to be found out. Concept Introduction: Uncertainty with a Calibration Curve: Propagation for the equation y = m x + b (but not y = m x ) gives, The standard uncertainty in x is given as: u x = s y | m | 1 k + 1 n + ( y − y ¯ ) 2 m 2 ∑ ( x i − x ¯ ) 2 Where, u x is standard uncertainty for x y is the corrected absorbance of unknown x i is the mass of sample in standards y ¯ is the average of y values x ¯ is the average of x values k is the number of replicate measurements

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Answer 1

Question

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

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Answer 2

Question

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

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Answer 3

Question

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

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Answer 4

Question

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

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Answer 5

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

Answer 6

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

Answer 7

Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$ , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

Answer 8

(a)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0±0.3$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|11+14+(2.58−3.5)2(0.61538)2(13.0) =0.38 =0.3 (rounded off)$

Therefore, the value for x is $2.0±0.3$

Answer 13

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0±0.3$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

Answer 15

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$ , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2$

Where,

$ux$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$xi$ is the mass of sample in standards

$y¯$ is the average of $y$ values

$x¯$ is the average of $x$ values

k is the number of replicate measurements

Answer 16

(b)

Expert Solution

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is $2.0±0.2$

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b⇒y=0.61538x+1.34615$

On rearranging, we get,

$x=y−bm =2.58−1.350.615 =2.00$

The value of $y¯$ and $x¯$ is calculated as follows,

$y¯=(2+3+4+5)/4 =3.5x¯=(1+3+4+6)/4 =3.5$

The $∑(xi−x¯)2$ is found as follows,

$∑(xi−x¯)2=(1−3.5)2+(3−3.5)2+(4−3.5)2+(6−3.5)2 =13.0$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$ux=sy|m|1k+1n+(y−y¯)2m2∑(xi−x¯)2 =0.19612|0.61538|14+14+(2.58−3.5)2(0.61538)2(13.0) =0.26 =0.2 (rounded off)$

Therefore, the value for x is $2.0±0.2$

Answer 21

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0±0.2$

Answer 22

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

Answer 23

(c)

Interpretation Introduction

Interpretation:

The $95%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$Confidence interval = x¯±tsn =x¯±tux (since standard uncertainty(ux)=s/n)$

Where,

$x¯$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$ux$ is standard uncertainty

To Find: The $95%$ confidence intervals for (a) and (b)

Answer 24

(c)

Expert Solution

Answer to Problem 4.33P

The $95%$ confidence intervals for (a) is $2.0±0.8$

The $95%$ confidence intervals for (b) is $2.0±0.6$

Answer 25

(c)

Expert Solution

Answer 26

(c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00±0.38$

The results of the measurement in part (b) is $2.00±0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90%$ confidence level for three degrees of freedom is $t90%=2.353$

The $90%$ confidence interval for part (a) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.38) =2.00(±0.89) =2.0(±0.8) (rounded off)$

The $90%$ confidence interval for part (b) is calculated as follows,

$confidence interval = x¯±tux90% confidence interval =2.00(±2.353×0.26) =2.00(±0.61) =2.0(±0.6) (rounded off)$

Answer 29

Conclusion

The $95%$ confidence intervals for (a) is calculated as $2.0±0.8$

The $95%$ confidence intervals for (b) is calculated as $2.0±0.6$

Answer 30

Ch. 4.1 - Prob. 1TY Ch. 4.1 - Prob. 2TY Ch. 4.1 - Prob. 3TY Ch. 4.2 - Prob. 4TY Ch. 4.3 - Prob. 5TY Ch. 4.4 - Prob. 6TY Ch. 4.7 - Prob. 7TY Ch. 4.7 - Prob. 8TY Ch. 4.8 - Prob. 9TY Ch. 4 - Prob. 4.AE

Answer to Problem 4.33P

Explanation of Solution

Answer to Problem 4.33P

Explanation of Solution

Answer to Problem 4.33P

Explanation of Solution

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