A manufacture claim that their widget is more reliable than their main competitors. In order to verify this a sample of 65 widget from manufacturer’s range was taken. This mean rate found to be 992 per 1000 with sd 15 per 1000. previous studies have shown that the mean pass rate of all weight on the market is 979 per 1000. Test the manufacturer’s claim at 5% level of significance.

Solution:

Sample Size (n) = 65

Sample mean (x̄) = 992

Sample standard deviation(σ) = 15

Population mean(μ) = 979

Level of significance(α) = 5% = 0.05

Problem to Test:

H0: Average pass is 979 (μ = 979)

H1: Average pass is more than 979 (μ > 979)

Test Statistics:

$$z = \frac{\overline{X} – μ}{\frac{σ}{\sqrt{n}}}$$

$$= \frac{992 – 979}{\frac{15}{\sqrt{65}}}$$

= 6.987

Critical Value:

At α = 5% = 0.05, the critical value (Zα) = Z0.05 = 1.645

Decision:

Here, |z| = 6.987 > Ztab = 1.645, reject H0 at 5% level of significance.

Conclusion:

The claim of manufacture is correct.