A prosthetics company claimed that it created a new type of arm prosthetic that

Question

A prosthetics company claimed that it created a new type of arm prosthetic that has an average lifetime of at least 1650 days. The lifetime is approximately normally distributed with variance of 245. A consumer group designed a test to decide if there is evidence to refute the company's claim. They chose a significance level of alpha = 0.05 and randomly selected 32 prostheses to evaluate. a) Describe the null hypothesis and the alternative hypothesis. b) If the true mean found by the consumer group is mu = 1644, what is the power of this test? c) What can the consumer group do to increase the power of the test? d) What sample size should the consumer group utilize to achieve a power of 0.9 while maintaining the same significance level of alpha = 0.05 if the true mean is 1644 rather than 1650?

Explanation / Answer

a.
Null, H0: >=1650
Alternate, H1: <1650
b.
Given that,
Standard deviation, =15.6524758424985
Sample Mean, X =220
Level of significance, = 0.05
From Standard normal table, Z /2 =1.64
Since our test is left-tailed
Reject Ho, if Zo < -1.64 OR if Zo > 1.64
Reject Ho if (x-1650)/15.6524758424985/(n) < -1.64 OR if (x-1650)/15.6524758424985/(n) > 1.64
Reject Ho if x < 1650-25.6701/(n) OR if x > 1650-25.6701/(n)
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Suppose the size of the sample is n = 32 then the critical region
becomes,
Reject Ho if x < 1650-25.6701/(32) OR if x > 1650+25.6701/(32)
Reject Ho if x < 1645.4621 OR if x > 1654.5379
Implies, don't reject Ho if 1645.4621 x 1654.5379
Suppose the true mean is 1644
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1645.4621 x 1654.5379 | 1 = 1644)
= P(1645.4621-1644/15.6524758424985/(32) x - / /n 1654.5379-1644/15.6524758424985/(32)
= P(0.5284 Z 3.8084 )
= P( Z 3.8084) - P( Z 0.5284)
= 0.9999 - 0.7014 [ Using Z Table ]
= 0.2985
For n =32 the probability of Type II error is 0.2985

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