Question

Question 3 (7 marks) A manufacturer has developed a new type of bicycle frame which will be sold with a 2-year warranty. To see whether this is economically feasible, 20 prototype frames are subjected to an accelerated life experiment to simulate 2 years of use. The proposed warranty will be modified only if fewer than 90% of such frames would survive the 2-year period. (a) Let p be the true proportion of frames that survive. Find a rejection region for testing of the null hypothesis Ho : p = 0.9 against Hi: p < 0.9 at the 0.05 level of significance. Since the sample is relatively small, avoid normal approximations. (b) It is found that 14 frames survive the accelerated life test. Do these data suggest that the true proportion of for the frames that survive is smaller than 90%? (c) Find the probability of a Type II error assuming that the true proportion of frames surviving the accelerated life experiment is 0.8. Briefly comment on the accuracy of this test. How could you improve this result? You may use the following R output: > round(pbinom (10:20, size=20, prob=0.8), digit=2) [1] 0.00 0.01 0.03 0.09 0.20 0.37 0.59 0.79 0.93 0.99 1.00 > round (pbinom (10:20, size=20, prob=0.9), digit=2) [1] 0.00 0.00 0.00 0.00 0.01 0.04 0.13 0.32 0.61 0.88 1.00

Answer 1

solution:

a).

Let be the number of frames survive the 2-year period. Then X~ Binomial(n = 20, p = 0.9) if null hypothesis is true.

We reject null hypothesis Ho if,

p(X < x | p 0.9) < 0.05

From the R output, the largest value of x for which p(X < x | p = 0.9) < 0.05 is x = 15

P(X $\le$ 15 | p = 0.9) = 0.04

The rejection region is, X $\le$ 15. That is we reject H0 if number of frames survive the 2-year period is less than or equal to 15.

b).

Since the number of frames survive the 2-year period is less than 15, it lies in the rejection region and thus the data suggest that the proportion of for the frames that survive is smaller than 90%.

c).

Probability of Type II error = P(Fail to Reject Ho | Ho is false ) = p(False to Reject Ho | p = 0.8)

Probability of Type II error = P(Fail to Reject H0 | H0 is false) = P(Fail to Reject H0 | p = 0.8)

= P(X > 15 | p = 0.8)

= 1 - P(X $\le$ 15 | p = 0.8)

= 1 - 0.37 (From R output)

= 0.63

Since the probability of Type II error is high, the test has low accuracy. We can improve the result by increasing the sample size from 20 to a large number (say 100).

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