12.76 Physical Sciences A new solar collector is being tested for use in chargin
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Question
12.76 Physical Sciences A new solar collector is being tested for use in charging batteries that can provide electricity for an entire home. A random sample of days was selected and the amount of solar radiation was measured (x, in langleys) for each. The total battery charge was measured as a proportion (y, between 0 and 1). The summary statistics are given below.
B0 = 0.2007 n = 21 MSE = 0.06135
a. What proportion of a charge can you expect the batteries to take if the amount of solar radiation is 100 langleys?
b. Find a 95% confidence interval for the true mean battery charge proportion if the amount of solar radiation is 130 langleys.
c. A value of 80 langleys indicates a typical cloudy day. On a typical cloudy day, is there any evidence to suggest that the true mean charge proportion is greater than 0.06 (the proportion needed to ensure a home will have sufficient energy until the next day)? Use a significance level of 0.01.
Explanation / Answer
a. Substituting the values of y intercept->beta0, and slope->beta1 in the regression equation, yhat=beta0+beta1*x format, the following regression eqaution is obtained.
proportion of charge= 0.2007+0.00446*langleys
Now, substitute langleys with 100 in the preceding equation and compute the proportion of charge.
Proportion of charge=0.2007+0.00446*100
=0.6467 (ans)
b. The 95% c.i for the true mean battery charge proportion for x=130 is computed as follows:
yphat=0.2007+0.00446*130=0.7805
for n=21, degrees of freedom=19 [df=n-2], t critical at alpha/2 is 2.09.
Therefore, the end points of the confidence interval for conditional mean of response variable is:
yphat+-talpha/2*Se*sqrt[1/n+{(xp-xbar)^2/Sxx}], where, Se=sqrt SSE/n-2=sqrt [(0.06135/19)/19]=0.013, where, SSE=MSE/n-2=0.06135/19
=0.7805+-2.09*0.013*sqrt[1/21+{(130-103.095)^2/12335.8}]
=(0.7716, 0.7894)
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