A researcher conducted a hypothesis test to see if students on average spend more than 1 hour per homework assignment. The level
of significance, a, selected was 0.05. This means that if on average they do spend exactly 1 hour per assignment, then there is a 5% chance that the erroneous conclusion will be made that students spend on average more than 1 hour per assignment. a. Trueb. False
By ∝= 5% we mean that there are about 5 chances in 100 of incorrectly rejecting a true null hypothesis. To put it in another way , we say that we are 95% confident in making the correct decision.
In the given question the null hypothesis is
H0: u ≤ 1 hour and Ha: u > 1 hour
So there is a 5% chance that the erroneous conclusion will be made that students spend on average more than 1 hour per assignment.
We know that if we had 8/6=2/2 times 4/3=1 times 4/3=4/3 find common factors in top and obttom factor 48a^4-16a^2-32=(16)(a-1)(a+1)(3a^2+2) 8a^2-8=8(a-1)(a+1) so we have (2)(3a^2+2)=6a^2+4