Question 3. Researchers examined the time in minutes before an insulating fluid

Question

Question 3. Researchers examined the time in minutes before an insulating fluid lost its insulating property. The following data are the breakdown times for eight samples of the fluid, which had been randomly allocated to receive one of to voltages of electricity: Times (min) at 26 kV: 5.79 1579.52 2323.70 Times (min) at 28 kV: 68.8 108.29 110.29 426.07 1067.60 (a) Form two new variables by taking the logarithms of the breakdown times: Yi-log(breakdown time at 26kV) and Y2 = log(breakdown time at 28kV). (b) By hand, compute the difference in averages of the log-transformed data: Y - Y2. (c) Take the antilogarithm of the estimate in (b): epi- what does this estimate? (See the 9 interpretation for the randomized experiment model in Section 3.5.2 of The Statistical Sleuth or refer to the lecture note). (d) By hand, compute 95% confidence interval for the difference in mean log breakdown the reverse the log-transformation of the endpoints and express the result in times. Take a sentence. Assume equal population standard deviations. Question 4. Suppose we want to test if the annual defection rate of a machine part working under extreme conditions has recently depreciated: Ho : p = 0.1 vs Ha : p > 0.1, under 0.10 Suppose we plan to collect a sample of size n = 50. (a) What is the power of the test, if the tru e annual defection rate now is p 0.15? (b) If we want to achieve a power of at least 95%, what is the least sample size needed?

Explanation / Answer

(b)

mean of Y1=sum(Y1)/3=16.8717/3=5.6239

mean of Y2=sum(Y2)=sum(Y2)/5=26.6469/5=5.3294

Y1--Y2-=5.6239-5.3294=0.2945

(c) antillograrithm of (Y1--Y2-)=antilog(0.2945)=exp(0.2945)=1.3425

(d)

pooled variance= sp2=((n1-1)s12+(n2-1)s22)/n and n=n1+n2-2

sp2=((3-1)*11.2571+(5-1)*1.3104)/(3+5-2)=4.6259

sp=sqrt(4.6259)=2.1508

95% confidence interval for difference of mean=difference of mean±t(.05, 6)*sp/sqrt(n) (here df for t= n=6)

=0.2945±2.4469*2.1508/sqrt(6)=0.2945±2.1485=(-1.8540,2.4430)

26kV 28kV Y1=log(26kV) Y2=Log(28kV) 5.79 68.8 1.756132292 4.231203745 1579.2 108.29 7.364673669 4.684812814 2323.7 110.29 7.750916022 4.70311326 426.07 6.054603652 1067.6 6.973168418

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.