A new electric car has been clocked in Japan to have a range of 71.9km. An indep

Question

A new electric car has been clocked in Japan to have a range of 71.9km. An independent test in Trinidad and Tobago consisting of 27 trials produced the following summary statistics in Excel Ranges (km) Mean SE of mean Median Mode Standard Deviation Sample Variance Range Minimum Maximum Sum Count 72.699 72.984 74.673 2.265 5.132 9.978 66.144 76.122 2180.978 27 a. Identify the biased estimate(s) for the population mean range of electric vehicles in Trinidad and Tobago. Identify the estimate for the population variance. Fill in the gap marked by. 12) 13] iv. Assuming normality, state clearly the estimated parameters of the distribution of the 12) [5] ii. range of the electric cars. Calculate the probability that the range of an electric car will be at least 60km. Calculate a 90% confidence interval for the mean range of the vehicles and give an interpretation of the results. V. vi. b. i The dealer for the electric cars would like to determine whether the mean range of the cars is greater than the manufacturer's stated range. Give the alternative hypothesis for the test. i. State Ho. iii. Give the rejection region for the test using a significance level of 5%. Conduct the hypothesis test and state your conclusions. Give an approximate p-value for the test, and interpret it. 141 V.

Explanation / Answer

(a )(i) biased=difference(71.9-72.699)=0.799

(ii)population variance=2.265*2.265=5.1302

(iii)SE of mean=sd/sqrt(n)=2.265/sqrt(27)=0.4359

(iv)Normal(72.699,5.1302), mean=72.699 and variance=5.1302

(v)P(X>60)=P(Z>(60-72.699)/2.265)=P(Z>-5.6)=1-P(Z<-5.6)=1-0=1 ,

here we use standard normal variate z=(x-mean)/sd, for x=60, z=-5.6

that is almost all will have range atleast 60

(vi) (1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

90% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)

(b)

(b)(i) alternate hypothesis Ha:mu>71.9

One-tailed test

(ii)null hypothesis H0:mu<=71.9

(iii) rejection region Z>Z(0.05)1.645

This is one tailed rejection region

(iv) z=(72.699-71.9)/(2.265/sqrt(27))=1.83 is more than critical z=1.645, we fail to accept H0 and conclude that range of the car is greater than the manufacturer started range

(v) p-value=0.0336 ( P(Z>1.83))

n= 27 sample mean= 72.699 sd= 2.27 z-value lower limit upper limit 90% confidence interval 1.65 71.98 73.42

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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