Question

A machine that is programmed to package 2.65 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 36 cereal boxes, the mean and the standard deviation are calculated as 1.22 pounds and 0.06 pound, respectively. b-1. Calculate the value of the test statistic.

Answer 1

To determine the test statistic, use the formula z = (x - μ) / (σ/√n), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plug in the given values and solve to find the value of the test statistic.

Step-by-step explanation:

To determine whether the machine needs to be recalibrated, we need to calculate the test statistic, also known as the z-score. The formula for calculating the z-score is:

z = (x - μ) / (σ/√n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this case, the sample mean is 1.22 pounds, the population mean is 2.65 pounds, the population standard deviation is 0.06 pound, and the sample size is 36. Plugging these values into the formula, we get:

z = (1.22 - 2.65) / (0.06 / √36)

Simplifying the expression, we get:

z = -1.43 / 0.01

Therefore, the value of the test statistic is -143.

Answer 2

Using the z-score formula, the value of the test statistic is -143.

Z-score Formula:

z = (
`\bar x` - μ) / (σ / √n)

Where:

The sample mean =
`\bar x`

The population mean = μ

The population standard deviation = σ

The sample size = n

The mean quantity of cereal to be packaged in each cereal box = 2.65 pounds

The number of cereal boxes sampled = 36

The sample mean = 1.22 pounds

The standard deviation = 0.06 pound

z = (1.22 - 2.65) / (0.06 / √36)

= (-1.43) / (0.06 / 6)

= -1.43 / 0.01

= -143

Thus, the value of the test statistic (z) is -143.