An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students show that only 44 students have ever done so. Is there enough evidence to show that the true proportion is lower than 60%? Assume that significance level of 0.05. Use the hypothesis testing template provided.'

We accept H₀ we don´t have enough evidence to conclude that a consumer group position is correct

Step-by-step explanation:

We have a case of test of proportion, as a consumer group is suspicious of the claim and think the proportion is lower we must develop a one tail test (left tail) Then

1.- Test hypothesis:

Null hypothesis  H₀                   P = P₀

Alternative hypothesis  Hₐ       P < P₀

2.- At significance level of α  = 0,05   Critical value

z(c)  =  -1,64

3.-We compute z(s) value as:

z(s)  =  ( P - P₀ )/ √P*Q/n      where   P = 44/80     P = 0,55   and Q = 0,45

P₀ = 0,6   and  n = 80

Plugging all these values in the equation we get:

z(s)  = ( 0,55 - 0,6 ) / √(0,2475/80)

z(s)  =  - 0,05/ √0,0031

z(s)  =  - 0,05/0,056

z(s)  =  - 0,8928

4.-We compare  z(s)  and  z(c)

z(s) > z(c)      -0,8928 on the left side it means that z(s) is in the acceptance region so we accept H₀

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

can i see the following ?

step-by-step explanation: