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A newly electric vehicle is designed to have a battery range of 400 miles and a standard deviation of 20 miles. assume the battery range distribution is normally distributed. the company tests 39 vehicles to determine if the battery is operating at specifications. the average range of the 39 vehicles is 398 vehicles with a standard deviation of 20 miles. find the 95% confidence interval for average battery range of the newly produced vehicle.
a) (393.72 , 406.28)
b) (391.52, 404.48
c) (393.52, 406.48)
d) (391.72, 404.28)

Sagot :

Answer:

b) (391.52, 404.48)

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 39 - 1 = 38

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 38 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.0244

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.0244\frac{20}{\sqrt{39}} = 6.48[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 398 - 6.48 = 391.52.

The upper end of the interval is the sample mean added to M. So it is 398 + 6.48 = 404.48.

The CI is (391.52, 404.58), and the correct answer is given by option B.

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