Samples of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered 50 mph. The resulting 95% confidence intervals for true mean actual speed were (51.7, 53.1) and (49.1, 50.3). Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size

n₁  > n₂     Sample size of the first sample is bigger

Step-by-step explanation:

CI = 95 %    α = 5 %     α = 0,05

We know the bell shape curve of the normal distribution is symmetrical with respect to the mean, therefore we only need to evaluate one tail of the samples

Then from z-table we find   z(c)  =  1,64

CI   =  (  μ₀ - z(c)*σ /√n   <  X  <   μ₀ + z(c)*σ /√n )

So we must compare the tails of the two CI

Let´s call  n₁    and   n₂   the samples size of two different samples

The CI of the samples begins at

sample 1       μ₀ - z(c)*σ /√n₁        and for sample 2    μ₀ - z(c)*σ /√n₂      

that means   51,7  =  μ₀ - z(c)*σ /√n₁     (1)

and                49,1  =   μ₀ - z(c)*σ /√n₂     (2)

Where   μ₀ ;  z(c) = 1,64 ;  σ  are equal in both equations

Now by simple inspection, we note that the second term on the right side of the second equation should be bigger than the second term on the right side of  the first equation

then    

z(c)*σ /√n₂   >  z(c)*σ /√n₁

Then   n₂    need to be smaller than  n₁

n₁  > n₂

answer: can you me with homework?

step-by-step explanation:

the solution is h=10 because you add 9 to each side of the equation.