Question

A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is known that the standard deviation of the population is 22 mph. The 95 percent confidence interval for μ is determined as (61.08, 68.92). If we are to reduce the sample size to 100 (other factors remain unchanged), the 95 percent confidence interval for μ would: Question 6 options: be the same. become wider. become narrower. be impossible to determine.

Answer 1

Answer:

95 percent confidence interval for μ is determined by

(60.688 ,  69.312)

a) be the same

Step-by-step explanation:

Step (i):-

Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.

The sample size is n= 121

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (ii) :-

Given  The 95 percent confidence interval for μ is determined as

(61.08, 68.92)

we are to reduce the sample size to 100 (other factors remain unchanged) so

given The sample size is n= 100

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (iii):-

95 percent confidence interval for μ is determined by

$(x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )$

$(65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )$

(65 - 4.312 ,65 +4.312)

(60.688 ,  69.312) ≅(61 ,69)

This is similar to (61.08, 68.92) ≅(61 ,69)

Conclusion:-

it become same as (61.08, 68.92)