Question

The output voltage for an electric circuit is specified to be 126. A sample of 40 independent readings on the voltage for this circuit gave a sample mean 124.2 and standard deviation 1.6.

(a)

Construct a 95% upper confidence bound for the average voltage reading. (Round your answer to three decimal places.)

How does the value μ = 126 compare to this upper bound?

The upper bound is  ---Select--- < = > 126.

(b)

Consider a test of

H₀: μ = 126

versus

H_a: μ < 126.

Based on the upper bound in part (a), should the alternative hypothesis of this test be accepted?

The value μ = 126 is below the upper bound found in part (a) providing sufficient evidence to conclude that the alternative hypothesis should be accepted.

The value μ = 126 is above the upper bound found in part (a) providing sufficient evidence to conclude that the alternative hypothesis should be accepted.    

The value μ = 126 is below the upper bound found in part (a) which does not provide sufficient evidence to conclude that the alternative hypothesis should be accepted.

The value μ = 126 is equal to the upper bound found in part (a) providing sufficient evidence to conclude that the alternative hypothesis should be accepted.

The value μ = 126 is above the upper bound found in part (a) which does not provide sufficient evidence to conclude that the alternative hypothesis should be accepted.

(c)

Suppose a test of

H₀: μ = 126

versus

H_a: μ < 126

is conducted with α = 0.05. The test rejected the null hypothesis. Is there any conflict between the answer in part (b) and the results of this test?

There is no conflict between the results as both suggest that the alternative hypothesis should be accepted.

There is a conflict between the results. The upper bound does not suggest the alternative hypothesis should be accepted while the hypothesis test suggests that the alternative hypothesis should be accepted.    

Answer 1

a)

sample mean, xbar = 124.2
sample standard deviation, s = 1.6
sample size, n = 40
degrees of freedom, df = n - 1 = 39

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.023

ME = tc * s/sqrt(n)
ME = 2.023 * 1.6/sqrt(40)
ME = 0.5

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (124.2 - 2.023 * 1.6/sqrt(40) , 124.2 + 2.023 * 1.6/sqrt(40))
CI = (123.688 , 124.712)

b)

The value μ = 126 is below the upper bound found in part (a) providing sufficient evidence to conclude that the alternative hypothesis should be accepted.

c)
There is no conflict between the results as both suggest that the alternative hypothesis should be accepted.