Four more I hoped this helped :)
In finding this value you average lower and upper bound
(0.6+0.82)/2 = 0.71
=0.71 estimated
margin of error = distance from estimate point lower/ upper bound
This interval will be twice margin error
(0.82-0.6)/2 = 0.11
How far is 0.82 from 0.71 ??
=0.11=11%
Answer:
the mode is 7 because they have the most points
Step-by-step explanation:
Answer:
We accept H₀ . We don´t have enough evidence to express the publisher claim is not true
Step by Step explanation:
We must evaluate if the mean of the price of college textbooks is different from the value claimed by the publisher
n < 30 then we must use t - distrbution
degree of freedom n - 1 df = 22 - 1 df = 21
As the question mentions " different " that means, a two-tail test
At 0,01 significance level α = 0,01 α/2 = 0,005
and t(c) = 2,831
Test Hypothesis
Null Hypothesis H₀ μ = μ₀
Alternative hypothesis Hₐ μ ≠ μ₀
To calculate t(s)
t(s) = ( μ - μ₀ ) /σ/√n
t(s) = ( 433,50 - 385 ) / 86,92 / √22
t(s) = 2,6171
Comparing t(c) and t(s)
t(s) < t(c)
Then t(s) is in the acceptance region we accept H₀. We don´t have enough evidence to claim that mean price differs from publisher claim
