Question: Failure times of silicon wafer microchips. Researchers at National Semiconductor experimented with tin-lead solder bumps used to manufacture silicon wafer integrated circuit chips (International Wafer-Level Packaging Conference, November 3–4, 2005). The failure times of the microchips (in hours) were determined at different solder temperatures (degrees Celsius). The data for one experiment are given in the table. The researchers want to predict failure time (y) based on solder temperature (x).

1. Construct a scatterplot for the data. What type of relationship, linear or curvilinear, appears to exist between failure time and solder temperature?
2. Fit the model,$\mathit{E}\mathbf{\left(}\mathit{y}\mathbf{\right)}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}$, to the data. Give the least-squares prediction equation.
3. Conduct a test to determine if there is upward curvature in the relationship between failure time and solder temperature. (use.$\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$)

Step by step solution

01

Scatterplot

Time to failure (Y)	Temperature(X)
200	165	27225
200	162	26244
1200	164	26896
500	158	24964
600	158	24964
750	159	25281
1200	156	24336
1500	157	24649
500	152	23104
500	147	21609
1100	149	22201
1150	149	22201
3500	142	20164
3600	142	20164
3650	143	20449
4200	133	17689
4800	132	17424
5000	132	17424
5200	134	17956
5400	134	17956
8300	125	15625
9700	123	15129

It appears that there is a curvilinear relationship between failure time and solder temperature. From the scatterplot, it appears that there is a negative relationship between failure time and solder temperature which is indicated by the downward-sloping curve.

02

Least squares prediction equation 

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.970315
R Square	0.941512
Adjusted R Square	0.935355
Standard Error	688.1366
Observations	22
ANOVA
	df	SS	MS	F	Significance F
Regression	2	144830279.6	72415140	152.9256	1.94E-12
Residual	19	8997106.744	473531.9
Total	21	153827386.4
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	154242.9	21868.47384	7.053209	1.03E-06	108471.7	200014.2	108471.7	200014.2
Temperature(X)	-1908.85	303.6635562	-6.28607	4.92E-06	-2544.43	-1273.28	-2544.43	-1273.28
	5.928945	1.04764072	5.659331	1.86E-05	3.736208	8.121683	3.736208	8.121683

From the excel output, the values for intercept and slope coefficient are calculated. Therefore, the least square prediction equation is$\hat{\mathbf{y}}\mathbf{=}\mathbf{154242}\mathbf{.}\mathbf{9}\mathbf{-}\mathbf{1908}\mathbf{.}\mathbf{85}\mathit{x}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{928945}{\mathit{x}}^{\mathbf{2}}$

03

Significance of β2

$$\begin{gathered} {\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0} \\ {\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{>}\mathbf{0} \end{gathered}$$

Here, t-test statistic

$\mathbf{=}\frac{\widehat{{\mathbf{\beta }}_{\mathbf{2}}}}{\mathbf{s}{\mathbf{\beta }}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{928945}}{\mathbf{1}\mathbf{.}\mathbf{047641}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{659}$

Value of${\mathit{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{22}}$is 1.717

is rejected if t statistic >${\mathit{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{385}}$. For $\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, since t >localid="1649842001962" ${\mathit{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{199}}$

Sufficient to reject at 95% confidence interval.

Therefore,${\mathit{\beta }}_{\mathbf{2}}\mathbf{>}\mathbf{0}$

This means that the parabola has an upward curvature.

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