Samples of two different types of automobiles were selected, and the actual speed for each car was determined when the speedometer registered 50 mph. the resulting 95% confidence intervals for true average actual speed were (51.7, 53.1) and (48.9, 50.1). assuming that the two sample standard deviations are identical, which confidence interval is based on the larger sample size?

The interval [48.9, 50.1] was based on the larger sample size.

Step-by-step explanation:

Hello!

To answer this I'll assume that both CI were constructed using statistic Z.

The formula is

                      x[bar]± * (δ/√n)

I'll not use the whole formula, just the semi amplitude (d) of the CI

                             d= * (δ/√n)

Then we rewrite the formula based on the sample size

                         d* =δ/√n

                         ⇒ (d* )² = δ²/n

                         ⇒ n*(d* )² = δ²

                         ⇒ n= δ²/(d*)²

First interval:

                         [51.7, 53.1]

Amplitude: a= Ls- Li = 53.1 - 51.7 = 1.4

Semi-amplitude: d=a/2 = 1.4/2 = 0.7

Replace the formula

n₁= δ²/(d*)²

⇒ n₁= δ²/(0.7*1.96)²

⇒ n₁= δ²/1.88

Second Interval

                         [48.9, 50.1]

Amplitude: a= Ls- Li = 50.1 - 48.9 = 1.2

Semi-amplitude: d=a/2 = 1.2/2 = 0.6

n₂= δ²/(d*)²

⇒ n₂= δ²/(0.6*1.96)²

⇒ n₂= δ²/1.38

Now, we compare the information we've got for the two intervals

n₁= δ²/1.88

n₂= δ²/1.38

See that I didn't put a subscript under the variance in both formulas, this is because the text says, both standard deviation are identical, meaning that the same happens with the variance. You can prove mathematically, that the formula with the lowest denominator will have the largest result.

>For example, let's say the standard deviation (δ) is 5, then:

n₁= 5²/1.88 = 13.29

n₂= 5²/1.38 = 18.12

The interval [48.9, 50.1] was based on the larger sample size.

I hope you have a SUPER day!

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