Answer:
a) -6 mpg.
b) 2.77 mpg
c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).
Step-by-step explanation:
To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Gas mileage A: Mean 36, standard deviation 6, sample of 50:
So
![\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485](https://tex.z-dn.net/?f=%5Cmu_A%20%3D%2036%2C%20s_A%20%3D%20%5Cfrac%7B6%7D%7B%5Csqrt%7B50%7D%7D%20%3D%200.8485)
Gas mileage B: Mean 42, standard deviation 8, sample of 50:
So
![\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314](https://tex.z-dn.net/?f=%5Cmu_B%20%3D%2042%2C%20s_B%20%3D%20%5Cfrac%7B8%7D%7B%5Csqrt%7B50%7D%7D%20%3D%201.1314)
Distribution of the difference:
Mean:
![\mu = \mu_A - \mu_B = 36 - 42 = -6](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cmu_A%20-%20%5Cmu_B%20%3D%2036%20-%2042%20%3D%20-6)
Standard error:
![s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_A%5E2%2Bs_B%5E2%7D%20%3D%20%5Csqrt%7B0.8485%5E2%2B1.1314%5E2%7D%20%3D%201.4142)
A. Find the point estimate.
This is the difference of means, that is, -6 mpg.
B. Find the margin of error
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such
![M = zs = 1.96*1.4142 = 2.77](https://tex.z-dn.net/?f=M%20%3D%20zs%20%3D%201.96%2A1.4142%20%3D%202.77)
The margin of error is of 2.77 mpg
C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)
The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg
The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg
The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).