The miles per gallon (mpg) of gasoline in city that owners of Toyota Prius 2017

Question

The miles per gallon (mpg) of gasoline in city that owners of Toyota Prius 2017 actually get is assumed to be normally distributed. You surveyed 20 randomly chosen Prius owners, and followings are the sample mpg data of their cars. 49.8 50.8 51.00 52.0 52.2 52.2 52.2 52.9 53.2 53.6 53.9 55.1 55.6 55.9 55.9 56.0 56.1 56.6 58.8 62.7 The sum of all 20 data sigma^n_i = 1 X_i is 1086.5 and the Sum of Squares sigma^n_i = 1 (X_i - X)^2 is 175.4 u) Now our interest shifts from the population mean to the population standard deviation. What is the best point estimate of population standard deviation, sigma? v) Find the Critical Value X^2 R and X^2 L for a 90% confidence interval? Degree of freedoms is still df = n-1 = 19 X^2 L = X^2R = w) What is the 90% confidence interval for sigma? x) Interpret the above answer. y) Find the Critical Values X^2 R and X^2 L for a 99% confidence interval? Degree of freedoms is same as (e). X^2 L = X^2 R = z) What is the 99% confidence interval for sigma?

Explanation / Answer

The best point estimate of population standard deviation ( ) = sample standard deviation (s)

s = [(xi - x)2 / n-1] =[ 175.4/19] = 3.03833

for 90% confidence interval

= 1-0.90 = 0.1

From chi-square table:

2/2,19 = 0.05,19 = 30.144

21-/2,19 = 0.95,19 = 10.117

Confidence interval is given by:

[(n-1)s2/ 2/2] [(n-1)s2/ 21-/2]

[19*9.231579/30.144] [19*9.231579/10.117]

2.412206 4.163791

We are 90% confident that the variation in population data lies between 2.412 and 4.164.

for 99% confidence interval

= 1-0.99 = 0.01

from chi-square table:

2/2,19 = 0.005,19 = 38.582

21-/2,19 = 0.995,19 = 6.844

Confidence interval is given by:

[(n-1)s2/ 2/2] [(n-1)s2/ 21-/2]

[19*9.231579/38.582] [19*9.231579/6.844]

2.1322 5.0624

We are 90% confident that the variation in population data lies between 2.1322 and 5.0624.

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