Question

A test engineer wants to estimate the mean gas mileage (in miles per gallon) for a particular model of automobile. Eleven of these cars are subjected to a road test, and the gas mileage is computed for each car. A dot plot of the 11 gas-mileage values is roughly symmetrical and has no outliers. The mean and standard deviation of these values are 25.5 and 3.01, respectively. Assuming that these 11 automobiles can be considered a simple random sample of cars of this model, which of the following is a correct statement?
a. A 95% confidence interval for μ is 25.5 ±2.2284 3.01
b. A 95% confidence interval for μ is 25.5±2.201 3.01
c. A 95% confidence interval for μ is 25.5±2.228 10 3.01
d. A 95% confidence interval for μ is 25.5±2.201 10
e.The results cannot be trusted; the sample is too small.

Answer 1

Answer:

a) A 95% confidence interval for μ is 25.5 ±2.2284 3.01

The 95% of confidence intervals for mean μ is determined by

(23.478 , 27.522)

Step-by-step explanation:

Step( i ) :-

Given sample size 'n' =11

The mean of the sample x⁻ = 25.5

The standard deviation of the sample 'S' = 3.01

95% of confidence intervals:

The 95% of confidence intervals for mean μ is determined by

$( x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )$

Step(ii):-

The critical value ∝ =0.05

$t_{\frac{\alpha }{2} } = 2.228$

The degrees of freedom ν=n-1 = 11-1 =10

$( 25.5 - 2.228 \frac{3.01}{\sqrt{11} } , 25.5 + 2.228 \frac{3.01}{\sqrt{11} } )$

(25.5-2.0220, 25.5 + 2.0220)

(23.478 , 27.522)

Final answer:-

The 95% of confidence intervals for mean μ is determined by

(23.478 , 27.522)