Homework Sourse1 A fleet of rental cars all the same make model and year
1. A fleet of rental cars - all the same make, model, and year - has a mean fuel efficiency of 27 miles per gallon (mpg). A random sample of 43 cars are selected and the air filter of each car is replaced with a new one. Let be the population mean fuel efficiency score that would occur if the air filter of every car in the fleet were replaced. The air filter change is deemed effective if µ > 27 mpg.
A test is made of H0: µ = 27 versus H1: µ > 27.
Assume that the air filter changes are NOT effective but the conclusion of the test was to reject the null hypothesis, that is the changes might be effective. Which type of error, if any, has occurred?
A) Type I B) Type II C) No Error – Correct decision D) Mechanical failure
2. A sample of 37 light bulbs had a mean lifetime of 584 hours. A 95% confidence interval for the population mean was 579.2 < < 588.8.
Which one of the following statements is the correct interpretation of the results?
A) The probability that the population mean is between 579.2 hours and 588.8 hours is 0.95.
B) We are 95% confident that the mean lifetime of all the bulbs in the population is between 579.2 hours and 588.8 hours.
C) 95% of the light bulbs in the sample had lifetimes between 579.2 hours and 588.8 hours.
D) We are 95% confident that most of the light bulbs had a life time between 579.2 hours and 588.8 hours.
3. In the US, adult male heights are normally distributed with mean 70 inches and standard deviation 2.5 inches. James is a professional basketball player and his height is in the 98th percentile. Which of the following is his height to the nearest inch?
A) 73 inches B) 80 inches C) 82 inches D) 75 inches
4. The proportion of registered voters that voted in the recent election was 55%. If samples of n = 200 registered voters are taken, which one is the standard deviation of the sampling distribution of the sample proportion?
A) 0.55 B) 3.89 C) 0.035 D) 0.001
Solution
Q3.
Mean ( u ) =70
Standard Deviation ( sd )=2.5
Normal Distribution = Z= X- u / sd ~ N(0,1)
P ( Z < x ) = 0.98
Value of z to the cumulative probability of 0.98 from normal table is 2.054
P( x-u/s.d < x - 70/2.5 ) = 0.98
That is, ( x - 70/2.5 ) = 2.05
--> x = 2.05 * 2.5 + 70 = 75.135 ~ D) 75 inches
Q4.
Proportion ( P ) =0.55
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.55*0.45/200) = 0.035
Note: Post the unanswered question in next thread
