21) A consumer group was interested in comparing the operating time of cordless toothbrushes manufactured by two different companies. Group members took a random sample of 18 toothbrushes from Company A and 15 from Company B. Each was charged overnight and the number of hours of use before needing to be recharged was recorded. Company A toothbrushes operated for an average of 119.7 hours with a standard deviation of 1.74 hours; Company B toothbrushes operated for an average of 120.6 hours with a standard deviation of 1.72 hours. The 90% confidence interval is (-1.93, 0.13). The correct interpretation is

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jtellezd jtellezd · Answer 1 · 2020-12-24T13:59:58+01:00

Answer:

We accept H₀ we don´t have evidence of differences between the two toothbrushes

Step-by-step explanation:

Toothbrushes A

Sample mean μₐ = 119,7

Standard deviation sₐ = 1,74

Sample size nₐ = 18

Toothbrushes B

Sample mean μ₀ = 120,6

Standard deviation s₀ = 1,72

Sample size n₀ = 15

With CI = 90 % CI = ( -1,93 ; 0,13 ) that s the acceptance region for H₀

Test hypothesis is

H₀ μₐ = μ₀

Hₐ μₐ ≠ μ₀

We need to calculate t(s) as

t(s) = ( μₐ - μ₀ ) /Sₓ * √ (1/nₐ) + 81/n₀)

and see if it is insde CI

Sₓ = (nₐ - 1 ) * sₐ² + ( n₀ - 1 ) * s₀² / nₐ + n₀ - 2

Sₓ = 17*3,03 + 14*2,96 / 31

Sₓ = (51,51 + 41,44 )/ 31

Sₓ = 2,998

t(s) = 0,9 / 2,998*√(1/18) + (1/15)

t(s) = 0,9 / 1,04

t(s) = 0,87