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, the mean ($\mu$) of jawbreakers weighing more than 0.4 ounces in a sample of 800 would be $800 \times 0.6 = 480$. The standard deviation ($\sigma$) for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \times p \times (1-p)}$, where $n$ is the sample size and $p$ is the probability of success. Substituting the given values, we get $\sigma = \sqrt{800 \times 0.6 \times 0.4} = \sqrt{192} \approx 13.86$. The range for significant values more than 2 standard deviations from the mean would be $\mu \pm 2\sigma$, which calculates to $480 \pm 2 \times 13.86$, or $452.28$ to $507.72$. Since 494 jawbreakers fall within this range, the result is not considered significant.

The Acme Candy Company Claims That 60% of the Jawbreakers