The Acme Candy Company claims that 60 % of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be significant for this sample of 800 to contain 494 jawbreakers that weigh more than 0.4 ounces? 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

The Acme Candy Company claims that  60 %  of the jawbreakers it produces weigh more than  0.4  ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be significant for this sample of 800 to contain 494 jawbreakers that weigh more than  0.4  ounces? 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 60% of the jawbreakers are expected to weigh more than 0.4 ounces, for a sample of 800, the expected number ($\mu$) of jawbreakers weighing more than 0.4 ounces is $800 \times 0.60 = 480$. The standard deviation ($\sigma$) for a binomial distribution is given by $\sqrt{n p (1-p)}$, where $n$ is the sample size and $p$ is the probability of success. Thus, $\sigma = \sqrt{800 \times 0.60 \times (1-0.60)} = \sqrt{800 \times 0.60 \times 0.40} = \sqrt{192} \approx 13.856$. The range for non-significant results, using the criterion of within 2 standard deviations from the mean, is $\mu \pm 2\sigma = 480 \pm 2 \times 13.856$, which is approximately $480 \pm 27.712$, or from about 452.288 to 507.712. Since 494 is within this range, the result is not considered significant.

The Acme Candy Company Claims That 60 μ−2σ\mu - 2 \sigmaμ−2σ

The Acme Candy Company Claims That 60 $\mu - 2 \sigma$