The claim is that the standard deviation of the speeds of drivers in Bradford is 8 mph. An experienced driver feels that this is very low. If a random speed check is conducted for 50 drivers, the standard deviation is 10.5 mph. Can we reject the null at

Null hypothesis is rejected. Standard deviation is > 8 miles per hour

Step-by-step explanation:

σ [Population standard deviation] = 8 , s [sample standard deviation] = s

n [no of items] = 8

H0 [Null] : σ = 8 ; H1 [Alternate - Right Tail] : σ > 8

χ2 = (n - 1) . s^2 / σ^2

= 49 x (10.5)^2 / 82 = 5402.25 / 64

χ2 = 84.410

df [degree of freedom] = n -1 = 50 - 1 = 49

P value (χ^2 49 > 84.410) = 0.00125

p = 0.0013

p < α ie 0.05

So, H0 is rejected

Hence we state that standard deviation is > 8 miles per hpur

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

what do u need to know

step-by-step explanation: