A quality control specialist at a pencil manufacturer pulls a random sample of 45 pencils from the assembly line. The pencils have a mean length of 17.9 cm. Given that the population standard deviation is 0.25 cm, find the P-value you would use to test the claim that the population mean of pencils produced in that factory have a mean length equal to 18.0 cm.

Question

A) 0.9926
B) 0.0074
C) 0.0037
D) 0.9963

Answer 1

To find the P-value for the hypothesis test regarding the mean length of pencils, we will conduct a hypothesis test using the following steps:

1. Define the null and alternative hypotheses:

   • Null hypothesis (H₀): μ = 18.0 cm (the population mean length of the pencils is 18.0 cm)
   • Alternative hypothesis (H₁): μ ≠ 18.0 cm (the population mean length of the pencils is not 18.0 cm)

2. Gather the sample data:

   • Sample mean (x̄) = 17.9 cm
   • Population standard deviation (σ) = 0.25 cm
   • Sample size (n) = 45

3. Calculate the standard error (SE): \[ SE = \frac{σ}{\sqrt{n}} = \frac{0.25}{\sqrt{45}} \approx \frac{0.25}{6.708} \approx 0.0372 \text{ cm} \]

4. Calculate the test statistic (z): \[ z = \frac{x̄ - μ}{SE} = \frac{17.9 - 18.0}{0.0372} \approx \frac{-0.1}{0.0372} \approx -2.688 \]

5. Determine the P-value: Since this is a two-tailed test, we need to find the probability of obtaining a z-value as extreme as -2.688 or as extreme as +2.688.

   Using the standard normal distribution table (or a calculator), we can find the P-value associated with z = -2.688. The area to the left of z = -2.688 gives approximately 0.0037.

   However, since it's a two-tailed test, we need to double this value to get the P-value: \[ P\text{-value} = 2 \times 0.0037 = 0.0074 \]

Thus, the P-value for the hypothesis test is 0.0074.

The answer is B) 0.0074.