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

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3