Question

Note: Express all values accurate to 4 decimals (0.1234) or as percentages accurate to 2 decimals (12.34%) unless otherwise stated.

You may work with others on this assignment.

(a: 1; b: 5; c: 2; d: 2; total: 10 marks)

In a manufacturing plant, dowel widths are normally distributed with an average of 5 mm and standard deviation of 0.02 mm. Each hour, a random sample of 25 dowels is taken to ensure that the average width is not significantly different from 5 mm. In one sample, the average was 5.0054 mm.

1. Why is the Z test the appropriate test for this situation?
2. Test the hypothesis at a 5% level of significance.
3. If the level of significance were 7.68%, what conclusion would be reached and why?
4. If a level of significance had not been specified, what conclusion would be reached and why?

Answer 1

(a) The Z test is the appropriate test for this situation because we know the population standard deviation.

(b) The hypothesis being tested is:

H0: µ = 5

Ha: µ ≠ 5

The test statistic, z = (x - µ)/σ/√n = (5.0054 - 5)/0.02/√25 = 1.35

The p-value is 0.1770.

Since the p-value (0.1770) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we can conclude that the average width is not significantly different from 5 mm.

(c) Since the p-value (0.1770) is greater than the significance level (0.0768), we cannot reject the null hypothesis.

Therefore, we can conclude that the average width is not significantly different from 5 mm.

(d) Since the p-value (0.1770) is greater than the significance level (0.10), we cannot reject the null hypothesis.

Therefore, we can conclude that the average width is not significantly different from 5 mm.