Question

A survey for brand recognition is done and it is determined that​ 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of​ 800, would it be significant to get 634 consumers who recognize the Dull Computer Company​ name? Consider as significant any result

Answer 1

Given Information:

Probability = p = 68% = 0.68

Population = n = 800

Answer:

it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.

Step-by-step explanation:

We can check whether it would be significant to get 635 consumers who recognize the Dull Computer Company by finding out the mean and standard deviation.

mean = μ = np

μ = 800*0.68

μ = 544

standard deviation = σ = √np(1-p)

σ = √800*0.68(1-0.68)

σ = 13.2 ≈ 13

we know that 99% of data fall within 3 standard deviations from the mean

μ ± 3σ = 544+3*13, 544-2*13

μ ± 3σ = 544+39, 544-39

μ ± 3σ = 583, 505

So we can say with 99% confidence that the number of consumers who can  recognize the name of Dull Computer Company will be from 505 to 583 and since 583 < 634 we can conclude that it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.

Answer 2

Answer:

It would be significant

Step-by-step explanation:

Population proportion of consumers who recognize the company name = 68% = 0.68

If 634 consumers out of the 800 randomly selected consumers recognize the company name, sample proportion = 634/800 = 0.7925.

It is significant to get 634 because the sample proportion of consumers who recognize the company name is greater than the population proportion.