Question

A plastics manufacturer has developed a new type of plastic trash can and proposes to sell them with an unconditional 6-year warranty. To see whether this is economically feasible, 20 prototype cans are subjected to an accelerated life test to simulate 6 years of use. The proposed warranty will be modified only if the sample data strongly suggests that fewer than 90% of such cans would survive the accelerated test. After the accelerated life test on the 20 prototype cans, 13 survive. Does this indicate that the proposed warranty should be modified? Use =0.05.

Answer 1

since -3.73 is less than 1.645, we reject H₀.

Therefore this indicate that the proposed warranty should be modified

Explanation:

Given that the data in the question;

p" = 13/20 = 0.65

Now the test hypothesis;

H₀ : p = 0.9

Hₐ : p < 0.9

Now lets determine the test statistic;

Z = (p" - p ) / √[p×(1-p)/n]

= (0.65 - 0.9) /√[0.9 × (1 - 0.9) / 20]

= -0.25 / √[0.9 × 0.1 / 20 ]

= -0.25 / √0.0045

= -0.25 / 0.067

= - 3.73

Now given that a = 0.05,

the critical value is Z(0.05) = 1.645 (form standard normal table)

Now since -3.73 is less than 1.645, we reject H₀.

Therefore this indicate that the proposed warranty should be modified