Question

The reputation of many businesses can be severely damaged due to a large number of defective items during shipment. suppose 300 batteries are randomly selected from a large shipment; each is tested and 9 defective batteries are found. at a 0.1 level of significance, does this provide evidence that the proportion of defective batteries is less than 5%? what is the p-value for the test.

Answer 1

P^ = 9/300 = 0.03
H0 = p < 5%
H1 = p > 5%

standard deviation of sample distribution = sqrt[p(1 - p) / n] = sqrt[0.05(1 - 0.05)/300] = sqrt(0.0001583) = 0.01258

test statistics, z = (p^ - p) / standard deviation = (0.03 - 0.05) / 0.01258 = -1.589
P(-1.589) = 1 - P(1.589) = 1 - 0.94402 = 0.05598

Since, p = 0.05598 < significant level of 0.1, we reject the H0.
i.e. There is no sufficient evidence to suggest that the proportion of defective batteries is less than 5%.

The p-value of the test is 0.05598