The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Find the p-value for a test to see whether the candy bars are smaller than they are supposed to be.
A)Between .05 and .10
B)Between .025 and .05
C)Between .01 and .025
D)Less than .01

Question

Quizplus · Accepted Answer

To find the p-value for a one-tailed test (since we're testing if the candy bars are smaller than they are supposed to be), we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{(\sigma/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we get $z = \frac{55.82 - 56}{(0.77/\sqrt{49})} = \frac{-0.18}{0.11} = -1.636$. Looking up the z-score of -1.636 in a standard normal distribution table or using a calculator, we find that the p-value is between .05 and .10, indicating the probability of observing a sample mean of 55.82 gm or less by chance if the true mean is 56 gm.

The Process That Produces Sonora Bars (A Type of Candy)