The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 24 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period.a. Which form of the null and alternative hypotheses most appropriate for this situation.
b. Comment on the conclusion when H0 cannot be rejected.
c. Comment on the conclusion when H0 can be rejected.

Answers

Answer 1
Answer:

Answer:

A) H_o ; μ ≤ 24

H_o ; μ > 24

B) There is no sufficient evidence to support the claim that the new bonus plan increases sales volume.

C) There is sufficient evidence to support the claim that the new bonus plan increases sales volume.

Step-by-step explanation:

A) In the question, the claim is to test whether the new bonus plan increases sales volume.

The claim is either the null hypothesis or the alternative hypothesis. The null and alternative hypothesis state the opposite of each other. Thus, the hypothesis for this problem is;

H_o ; μ ≤ 24

H_o ; μ > 24

B) If H_o cannot be rejected, then we can conclude that; there is no sufficient evidence to support the claim that the new bonus plan increases sales volume.

C) If H_o can be rejected, then we can conclude that; there is sufficient evidence to support the claim that the new bonus plan increases sales volume.


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Answers

Answer:

Y=-1/5x-1

Step-by-step explanation:

Combine the slope then the slope intercept.

The number of times a player has golfed in one's lifetime is compared to the number of strokes it takes the player to complete 18 holes. The correlation coefficient relating the two variables is -0.26. Which best describes the strength of the correlation and what is true about the causation between the variables?

Answers

Answer: It is a weak negative correlation and it is not likely causal.

Step-by-step explanation:

Given: The number of times a player has golfed in one's lifetime is compared to the number of strokes it takes the player to complete 18 holes. The correlation coefficient relating the two variables is -0.26.

Variables : "number of times a player has golfed in one's lifetime" and "number of strokes it takes the player to complete 18 holes".

Since -0.26 is more closer to 0 as compared to 1 , so it describes a weak negative correlation.

Also, it is not likely causal as number of times a player has golfed in one's lifetime not cause number of strokes it takes the player to complete 18 holes.

Answer: B) It is a weak negative correlation, and it is likely casual

Correct on edge 2020!

Show that f(x)=12x^3-32x^2+17 has a zero between 2 and 3.

Answers

Answer:

Step-by-step explanation:

You could graph it. I'll do that at the bottom. What you need is a value that is - on either 2 or 3 and a plus on the other one.

y=12*2^3 - 32x^2 + 17

y = 12 * 8 - 32*2^2 + 17

y = 96 - 128 + 17

y = -15

Now when x = 3 should give you something above the x axis

y = 12*3^3 - 32*9 + 17

y = 324 - 288 + 17

y = 53

What this indicates is that somewhere between 2 and 3 the graph crosses the x axis. The graph should indication that.

What is the following sum?
5(3squareroot x) +9 (3squareroot x)

Answers

\sf{14(\sqrt[3]{x}) }

Step-by-step explanation:

5(\sqrt[3]{x})+9(\sqrt[3]{x})\n\n(5+9)(\sqrt[3]{x})\n\n14(\sqrt[3]{x})

A professor graded the final exams and found that the mean score was 70 points. Which of the following can you conclude?A- All of the above.

B- The median score was 70 points.

C- 50% of the students scored below 70 points.

D- This would be a normal distribution.

Answers

Answer:  C) 50% of the students scored below 70%

Step-by-step explanation:

Mean is the average.  To find the mean (aka average) you add up all of the scores and divide by the number of tests.  

B) The mean can be 70 without any test scoring 70% so B is not true.

A) Since B is not true, then A is not a valid option.

D) We don't know any of the other data so don't know if it is skewed left, skewed right, or normal.  Therefore, option D is not true.

C) If the average is 70%, then half received grades above that score and half received grades below that score.  So, option C is TRUE!

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence? Assume that σ is known to be 0.72. (Exact answer required.)

Answers

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = (1-0.99)/(2) = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*(\sigma)/(√(n))

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when M = 0.08

We have that \sigma = 0.72

M = z*(\sigma)/(√(n))

0.08 = 2.575*(0.72)/(√(n))

0.08√(n) = 2.575*0.72

√(n) = (2.575*0.72)/(0.08)

(√(n))^(2) = ((2.575*0.72)/(0.08))^(2)

n = 537.1

Rounding up

538 books should be tested.