An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consumer group is

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.’

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Kylie 1 week 2021-11-20T19:41:01+00:00 1 Answer 0 views 0

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    2021-11-20T19:42:40+00:00

    Answer:

    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

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45:7+7-4:2-5:5*4+35:2 =? ( )