Measurements of the electrical resistivity of tungsten at specified temperatures are presented in the table below. The errorin the determination of the temperature was negligible. The temperatures are stated as absolute temperatures in °K. Theunits of resistivity are u Ncm .Carry out a regression analysis for resistivity as a function of temperature.Temperature Resistivity3005.54Include a scatterplot showing the residuals vs temperature.4007.95Do the regression assumptions appear to be valid?50011.0060012.28Include a scatterplot showing the raw data points as black circles as well as acontinuous blue line representing the regression model.70015.9580018.46Find the lower and upper limits for a 99% confidence interval for the slope of theregression model. Add a pair of red lines to the scatterplot: both passing through thecentroid of the data, one having the slope corresponding to the lower bound of theconfidence interval, and the other having the slope corresponding to the upper boundof the confidence interval.90020.47100023.50110026.71120028.65130032.10Add vertical error bars to the blue line representing the regression model representingthe top and bottom of the 99% confidence interval of the regression mean response foreach input temperature. You will need the two-tailed t-statistic with n-2 degrees offreedom, the standard error of the regression, Sy, and the sum of the squaredtemperature deviations, Sxx. Recall that the confidence interval is given by140034.82150037.40160040.811170042.36´known— тх + btcrityS.180045.79XX190047.97Compare the two ways of bounding the regression model: using the confidenceinterval for the slope, and using the confidence interval for the mean responses.200051.56If a new measurement will be made at a temperature of 300°K, what is the 99% prediction interval for that newmeasurement?

Name: Measurements of the electrical resistivity of tungsten at specified t
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Description: Measurements of the electrical resistivity of tungsten at specified temperatures are presented in the table below. The errorin the determination of the temperature was negligible. The temperatures are stated as absolute temperatures in °K. Theunits

As per our guidelines , we are allowed to answer three sub-parts only The following data are is given: Temperature Resistivity 300 5.54 400 7.94 500 11 600 12.28 700 15.95 800 18.46 900 20.47 1000 23.5 1100 26.71 1200 28.65 1300 32.1 1400 34.82 1500 37.4 1600 40.81 1700 42.36 1800 45.79 1900 47.97 2000 51.56 The independent variable is TTT, and the dependent variable is RRR. In order to compute the regression coefficients, the following table needs to be used: T R T*R T2 R2 300 5.54 1662 90000 30.6916 400 7.94 3176 160000 63.0436 500 11 5500 250000 121 600 12.28 7368 360000 150.7984 700 15.95 11165 490000 254.4025 800 18.46 14768 640000 340.7716 900 20.47 18423 810000 419.0209 1000 23.5 23500 1000000 552.25 1100 26.71 29381 1210000 713.4241 1200 28.65 34380 1440000 820.8225 1300 32.1 41730 1690000 1030.41 1400 34.82 48748 1960000 1212.4324 1500 37.4 56100 2250000 1398.76 1600 40.81 65296 2560000 1665.4561 1700 42.36 72012 2890000 1794.3696 1800 45.79 82422 3240000 2096.7241 1900 47.97 91143 3610000 2301.1209 2000 51.56 103120 4000000 2658.4336 Sum = 20700 503.31 709894 28650000 17623.9319Based on the above table, the following is calculated: Xˉ=1n∑i=1nXi=2070018=1150\bar X = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{ 20700}{ 18} = 1150 Yˉ=1n∑i=1nYi=503.3118=27.961666666667\bar Y = \frac{1}{n} \sum_{i=1}^{n} Y_i = \frac{ 503.31}{ 18} = 27.961666666667 SSXX=∑i=1nXi2−1n(∑i=1nXi)2=28650000−207002/18=4845000SS_{XX} = \sum_{i=1}^{n} X_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right)^2 = 28650000 - 20700^2/18 = 4845000SSYY=∑i=1nYi2−1n(∑i=1nYi)2=17623.9319−503.312/18=3550.54545SS_{YY} = \sum_{i=1}^{n} Y_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} Y_i\right)^2 = 17623.9319 - 503.31^2/18 = 3550.54545SSXY=∑i=1nXiYi−1n(∑i=1nXi)(∑i=1nYi)=709894−20700×503.31/18=131087.5SS_{XY} = \sum_{i=1}^{n} X_i Y_i - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right) \left(\sum_{i=1}^{n} Y_i\right) = 709894 - 20700 \times 503.31/18 = 131087.5 Therefore, based on the above calculations, the regression coefficients (the slope mm, and the y-intercept nn) are obtained as follows: m=SSXYSSXX=131087.54845000=0.0271m = \frac{SS_{XY}}{SS_{XX}} = \frac{ 131087.5}{ 4845000} = 0.0271 n=Yˉ−Xˉ⋅m=27.961666666667−1150×0.0271=−3.153n = \bar Y - \bar X \cdot m = 27.961666666667 - 1150 \times 0.0271 = -3.153 a) Therefore, we find that the regression equation is: R=−3.153+0.0271TR = -3.153 + 0.0271 T b) Scatter plot : c) Yes the Regression assumptions appears to be valid. As it is clearly visible from the graph that all the data points lie on the Regression line, which gives a very high correlation between Temperature and Resistivity There should be a linear and additive relationship between dependent (response) variable and independent (predictor) variable

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Statistics

Measurements of the electrical resistivity of tungsten at specified temperatures are presented in the table below. The error in the determination of the temperature was negligible. The temperatures are stated as absolute temperatures in °K. The units of resistivity are u Ncm . Carry out a regression analysis for resistivity as a function of temperature. Temperature Resistivity 300 5.54 Include a scatterplot showing the residuals vs temperature. 400 7.95 Do the regression assumptions appear to be valid? 500 11.00

Measurements of the electrical resistivity of tungsten at specified temperatures are presented in the table below. The error in the determination of the temperature was negligible. The temperatures are stated as absolute temperatures in °K. The units of resistivity are u Ncm . Carry out a regression analysis for resistivity as a function of temperature. Temperature Resistivity 300 5.54 Include a scatterplot showing the residuals vs temperature. 400 7.95 Do the regression assumptions appear to be valid? 500 11.00

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Question

Measurements of the electrical resistivity of tungsten at specified temperatures are presented in the table below. The error
in the determination of the temperature was negligible. The temperatures are stated as absolute temperatures in °K. The
units of resistivity are u Ncm .
Carry out a regression analysis for resistivity as a function of temperature.
Temperature Resistivity
300
5.54
Include a scatterplot showing the residuals vs temperature.
400
7.95
Do the regression assumptions appear to be valid?
500
11.00
600
12.28
Include a scatterplot showing the raw data points as black circles as well as a
continuous blue line representing the regression model.
700
15.95
800
18.46
Find the lower and upper limits for a 99% confidence interval for the slope of the
regression model. Add a pair of red lines to the scatterplot: both passing through the
centroid of the data, one having the slope corresponding to the lower bound of the
confidence interval, and the other having the slope corresponding to the upper bound
of the confidence interval.
900
20.47
1000
23.50
1100
26.71
1200
28.65
1300
32.10
Add vertical error bars to the blue line representing the regression model representing
the top and bottom of the 99% confidence interval of the regression mean response for
each input temperature. You will need the two-tailed t-statistic with n-2 degrees of
freedom, the standard error of the regression, Sy, and the sum of the squared
temperature deviations, Sxx. Recall that the confidence interval is given by
1400
34.82
1500
37.40
1600
40.81
1
1700
42.36
´known
— тх + bt
crit
y
S.
1800
45.79
XX
1900
47.97
Compare the two ways of bounding the regression model: using the confidence
interval for the slope, and using the confidence interval for the mean responses.
2000
51.56
If a new measurement will be made at a temperature of 300°K, what is the 99% prediction interval for that new
measurement?

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