Answer:
Step-by-step explanation:
The question is incomplete. The complete question is
The table below contains the overall miles per gallon (MPG) of a type of vehicle. Complete parts a and b below.
28, 34, 28, 20, 21, 31, 28, 24, 34, 35 , 36, 26, 25, 20
a. Construct a 95% confidence interval estimate for the population mean MPG for this type of vehicle, assuming a normal distribution.
b. Choose the correct answer below.
A. We have 95% confidence that the mean MPG of this type of vehicle for the sample is contained in the interval.
B. We have 95 ℅ confidence that the population mean MPG of this type of vehicle is contained in the interval. This is the correct answer.
C.95 % of the sample data fall between the limits of the confidence interval.Your answer is not correct.
D. The mean MPG of this type of vehicle for 95?% of all samples of the same size is contained in the interval.
Solution:
a) Mean = (28 + 34 + 28 + 20 + 21 + 31 + 28 + 24 + 34 + 35 + 36 + 26 + 25 + 20)/14 = 27.86
Standard deviation = √(summation(x - mean)²/n
n = 14
Summation(x - mean)² = (28 - 27.86)^2 + (34 - 27.86)^2 + (28 - 27.86)^2 + (20 - 27.86)^2 + (21 - 27.86)^2+ (31 - 27.86)^2 + (28 - 27.86)^2 + (24 - 27.86)^2 + (34 - 27.86)^2 + (35 - 27.86)^2 + (36 - 27.86)^2 + (26 - 27.86)^2 + (25 - 27.86)^2 + (20 - 27.86)^2 = 399.7144
Standard deviation = √(399.7144/14) = 5.34
Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
s = sample standard deviation = 5.34
n = number of samples = 14
From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 14 - 1 = 13
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
z = 2.16
Margin of error = 2.16 × 5.34/√14
= 3.08
The confidence interval is 27.86 ± 3.08
b) B. We have 95 ℅ confidence that the population mean MPG of this type of vehicle is contained in the interval.
