Question

ch 17 - population sigma known (σ)
example 1: the mean length of certain construction lumber is supposed to be 8.5 ft. a random sample of 81
pieces of such lumbers gives a sample mean of 8.3 ft. a builder claims that the mean of the lumber is different
from 8.5 ft. does the data support the builders claim at α=0.05, if the population standard deviation is 1.2 ft?

Explanation:

Step1: State hypotheses

Let \(H_{0}:\mu = 8.5\) (null hypothesis, mean length is \(8.5\) ft) and \(H_{1}:\mu
eq8.5\) (alternative hypothesis, mean length is different from \(8.5\) ft)

Step2: Calculate test - statistic

The formula for the \(z\) - test statistic is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)
Given \(\bar{x} = 8.3\), \(\mu = 8.5\), \(\sigma=1.2\), \(n = 81\)

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Step3: Determine critical values

For a two - tailed test with \(\alpha = 0.05\), the critical values are \(z_{\alpha/2}=\pm1.96\)

Step4: Make a decision

Since \(-1.96<-1.5 < 1.96\) (the test statistic \(z=-1.5\) does not fall in the rejection region), we fail to reject the null hypothesis.

Answer:

The data does not support the builder's claim at \(\alpha = 0.05\)