Question

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.'

Answer 1

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

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