Question #53548

Suppose a new production system will be implemented if a hypothesis test supports the conclusion that the new system increases production. Current system mean = 400

The appropriate null and alternative hypotheses in this case are?

a. Ho: µ ≥ 400 Ha: µ < 400
b. Ho: µ ≤ 400 Ha: µ > 400
c. Ho: µ = 400 Ha: µ ≠ 400
d. Ho: µ = 0 Ha: µ ≠ 0
e. Ho: µ ≠ 400 Ha: µ = 400

The new system was monitored for 40 hours and showed a mean production rate of 410 units per hour. Based on prior experience, a population standard deviation of 50 units can be used. Assume a .05 level of significance.

What is the critical value of the test statistic?

a. z = 1.96
b. t = 1.26
c. z = 1.26
d. t = 1.685
e. z = 1.64

What is the hypothesis test conclusion?

a. Do not reject Ho.
b. The test is inconclusive.
c. Reject Ho.
d. Reject Ho.
e. Do not reject Ho.

The calculated value of the test statistic is

a. z = 1.64.
b. z = 1.96.
c. t = 1.685.
d. t = 1.26.
e. z = 1.26.

Expert's answer

Answer on Question #53548 – Math – Statistics and Probability

1. Suppose a new production system will be implemented if a hypothesis test supports the conclusion that the new system increases production. Current system mean = 400

The appropriate null and alternative hypotheses in this case are?

a. Ho: μ≥400\mu \geq 400 Ha: μ<400\mu < 400

b. Ho: μ≤400\mu \leq 400 Ha: μ>400\mu > 400

c. Ho: μ=400\mu = 400 Ha: μ≠400\mu \neq 400

d. Ho: μ=0\mu = 0 Ha: μ≠0\mu \neq 0

e. Ho: μ≠400\mu \neq 400 Ha: μ=400\mu = 400

Solution

The conclusion that the new system increases production means that the system mean is bigger than 400.

**Answer:** b. Ho: μ≤400\mu \leq 400 Ha: μ>400\mu > 400.

2. The new system was monitored for 40 hours and showed a mean production rate of 410 units per hour. Based on prior experience, a population standard deviation of 50 units can be used. Assume a .05 level of significance.

What is the critical value of the test statistic?

a. z=1.96z = 1.96

b. t=1.26t = 1.26

c. z=1.26z = 1.26

d. t=1.685t = 1.685

e. z=1.64z = 1.64

Solution

We know population standard deviation and sample size is bigger than 30. So, we use z-distribution.

The one tailed z-critical for a .05 level of significance is 1.64.

**Answer:** e. z=1.64z = 1.64.

3. What is the hypothesis test conclusion?

a. Do not reject Ho.

b. The test is inconclusive.

c. Reject Ho.

d. Reject Ho.

e. Do not reject Ho.

Solution

Test statistic z=1.26z = 1.26 is less than z-critical (1.64). Thus, we don't reject the null hypothesis at a .05 level of significance.

Answer: a. and e. Do not reject Ho.

4. The calculated value of the test statistic is

a. z=1.64z = 1.64

b. z=1.96z = 1.96

c. t=1.685t = 1.685

d. t=1.26t = 1.26

e. z=1.26z = 1.26

Solution

z=xˉ−μ0σn=410−4005040=1.26.z = \frac {\bar {x} - \mu_ {0}}{\frac {\sigma}{\sqrt {n}}} = \frac {410 - 400}{\frac {50}{\sqrt {40}}} = 1.26.


Answer: e. z=1.26z = 1.26

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