An electrical firm manufactures light bulbs that have a lifetime that is approxi

Question

An electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 900 hours and a standard deviation of 50 hours. A lifetime test of n = 25 samples resulted in the sample average of 910 hours. Assume the significance level of 0.05. (a) Test the hypothesis H_0: mu = 900 versus H_1: mu notequalto 900 using a p-value. (b) Calculate the power of the test if the true mean lifetime is 905. (c) What sample size would be required to detect a true mean lifetime as low as 899 hours if we wanted the power of the test to be at least 0.95?

Explanation / Answer

Q5

(a) Test statistic

Z = (xbar - H )/ (s/sqrt(n))

Z= (910 - 900)/ (50/ sqrt(25)) = 10/ 10 = 1

so p - value = Pr( Z >1) = 1 - 0.8413 = 0.1587

(b) Power off the test = 1 - Pr (Type II error)

Probability of type II error means fail to rejecting the null hypothesis even if it is true.

so we wil first find the rejection region here where we can reject the null hypothesis.

we will reject the null hypothesis outside the 95% confidence interval which is

95% CI = 900 +- 1.96 * (50/sqrt(25) = (880.4, 919.6)

so if true mean is 905 so we have to calculate the probability that we wil reject the null hypothesis.

Pr (xbar > 919.6 ; 905; 10)

Z - value - (919.6 - 905)/10 = 1.46

so Pr (xbar > 919.6 ; 905; 10) = 1 - 0.9279 = 0.072

Pr (Type II error) = 0.928

and power = 1 - 0.928 = 0.072

(c) Power = 0.95

so Pr( Type II error) = 0.05

That means if true mean life time is as low as 899 hours

then Pr ( type II error) = Pr(xbar < C; 899 ; 50/ sqrt(n))

where C is the value below which it is critical region and n is tha sample size.

so C = 900 - 1.96 * 50/ sqrt(n)

here Pr( Type II error) = 0.05

so Z - value = +-1.645

1.645 = (900 - 1.96 * 50/ sqrt(n) - 899)/  (50/ sqrt(n) )

82.25/ sqrt(n) = 1 - 98/sqrt(n)

180.25/ sqrt(n) =1

sqrt (n) = 180.25

n = 32,490

so for sample size of the above given n , the power of the test will be 0.95.

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