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seraphim [82]
3 years ago
7

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

er 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.'
Mathematics
1 answer:
Rasek [7]3 years ago
7 0

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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Step-by-step explanation:

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Idk what I should do
GREYUIT [131]

Answer:

c.

Step-by-step explanation:

All of the things are correct with this one.

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3 years ago
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Julli [10]
The answer I think is C
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3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B6%7D%20%7D%20%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%7D"
Kryger [21]

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x^2/3

Step-by-step explanation:

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