Question

A manufacturer claims that fewer than 6% of its fax machines are defective. In a random sample of 100 such fax machines, 5% are defective. Find the P-value for a test of the manufacturer's claim. Round to the nearest thousandth.

Answer 1

Solution :

The null and alternative hypotheses are as follows :

H​​​​​​0 : P = 6% = 0.06 i.e. The proportion of defective fax machine is 0.06.

H​​​​​​1 : P < 0.06 i.e. The proportion of defective fax machine is less than 0.06.

To test the hypothesis we shall use z-test for single proportion. The test statistic is given as follows :

Where, p̂ is sample proportion, P is hypothesized value of population proportion, Q = 1 - P and n is sample size.

Sample proportion of defective fax machine is,

p̂ = 5% = 0.05

P = 0.06, Q = 1 - 0.06 = 0.94 and n = 100

The value of the test statistic is -0.4211.

Since, our test is left-tailed test, therefore we shall obtain left-tailed p-value for the test statistic. The left-tailed p-value is given as follows :

p- value = P(Z < value of the test statistic)

p-value = P(Z < -0.4211)

Using "pnorm" function of R we get, P(Z < -0.4211) = 0.337

Hence, p-value = 0.337

The p-value for a test of the manufacturer's claim is 0.337.

Please rate the answer. Thank you.