Respuesta :
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 â„… 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 â„… confidence that the population mean MPG of this type of vehicle is contained in the interval.
