Question

A lumber company is making boards that are 2716.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A sample of 23 is made, and it is found that they have a mean of 2714.9 millimeters with a standard deviation of 12.0. A level of significance of 0.1 will be used to determine if the boards are either too long or too short. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

Answer 1

Solution:-

n = 23, Mean = 2714.9, S.D = 12.0

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 2716
Alternative hypothesis: u 2716

Note that these hypotheses constitute a two-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 2.50217

z = (x - u) / SE

z = - 0.44

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic less than -0.44 or greater than 0.44.

Thus, the P-value = 0.66.

Interpret results. Since the P-value (0.66) is greater than the significance level (0.10), we cannot reject the null hypothesis.