Question

# The life in hours of a battery is known to be approximately normally distributed with standard...

The life in hours of a battery is known to be approximately normally distributed with standard deviation σ = 1.5 hours. A random sample of 10 batteries has a mean life of ¯x = 50.5 hours. You want to test H0 : µ = 50 versus Ha : µ 6= 50.

(a) Find the test statistic and P-value.

(b) Can we reject the null hypothesis at the level α = 0.05?

(c) Compute a two-sided 95% confidence interval for the mean battery life.

(d) What sample size would be required to ensure that β does not exceed 0.10 if the true mean life is 48 hours?

x = 50.5, σ = 1.5, n = 10, µ = 50

a) Test statistic: z = (x-µ)/(σ/n^0.5) = (50.5-50)/(1.5/10^0.5) = 1.054

p-value (Using Excel function NORM.S.DIST(z,cumulative)) = NORM.S.DIST(1.054,TRUE) = 0.854

b) Since p-value is more than 0.05, we do not the null hypothesis and conclude that µ = 50.

c) 95% confidence interval:

z0.05/2 = 1.96

x ± z*(σ/n^0.5) = 50.5 ± 1.96*(1.5/10^0.5) = 50.5 ± 0.93 = 49.57 < µ < 51.43

d) b = 48, µ = 50, σ= 1.5, β = 01.0

z = (b-µ)/(σ/n^0.5) = (50-48)/(1.5/n^0.5)

P(z) = β =0.10

Using standard normal tables, z = 3.06

(50-48)/(1.5/n^0.5) = 3.06

1.5/n^0.5 = 2/3.06 = 0.65

n^0.5 = 1.5/0.65 = 2.31

n = 2.31^2 = 5.34

n = 5

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