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 that differs from the mean by more than 2 standard deviations. That is, significant values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$
A) Yes
B) No

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 that differs from the mean by more than 2 standard deviations. That is, significant values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$
A) Yes
B) No

Quizplus · Accepted Answer

Given that 68% of consumers have heard of Dull Computer Company, the expected number of consumers who recognize the company in a survey of 800 is $0.68 \times 800 = 544$. The standard deviation ($\sigma$) for a binomial distribution is calculated using the formula $\sigma = \sqrt{n p (1-p)}$, where $n$ is the sample size and $p$ is the probability of success. Here, $n = 800$ and $p = 0.68$, so $\sigma = \sqrt{800 \times 0.68 \times (1-0.68)} = \sqrt{800 \times 0.68 \times 0.32} \approx 14.8$. Two standard deviations from the mean is $2 \times 14.8 = 29.6$, so the range of significant values is less than $544 - 29.6$ or greater than $544 + 29.6$. Since 634 is significantly greater than $544 + 29.6$, getting 634 consumers who recognize the Dull Computer Company name would be considered significant.

A Survey for Brand Recognition Is Done and It Is μ−2σ\mu - 2 \sigmaμ−2σ

A Survey for Brand Recognition Is Done and It Is $\mu - 2 \sigma$