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The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 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. Develop the null and alternative hypotheses most appropriate for this situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

Short Answer

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a) \( H_0: \mu = 14 \), \( H_a: \mu > 14 \). b) Sales remain unchanged. c) Sales increase.

Step by step solution

01

Understanding the Null Hypothesis

The null hypothesis (H_0) is a statement of no effect or no difference. In this context, it implies that the new bonus plan does not increase the mean sales volume of automobiles per month. Thus, the null hypothesis is set as: \( H_0: \mu = 14 \) where \( \mu \) is the mean number of automobiles sold under the new bonus plan.
02

Formulating the Alternative Hypothesis

The alternative hypothesis (H_a) is a statement that contradicts the null hypothesis. It proposes that the new bonus plan increases the sales volume. Hence, the alternative hypothesis is: \( H_a: \mu > 14 \). This indicates that the mean sales volume under the new bonus plan is greater than 14 automobiles per month.
03

Conclusion When Null Hypothesis Cannot Be Rejected

If \( H_0 \) cannot be rejected, it suggests that the data does not provide strong enough evidence to conclude that the new bonus plan increases the sales volume of automobiles. Therefore, the mean sales volume remains statistically unchanged and is considered to be 14 automobiles per month.
04

Conclusion When Null Hypothesis Can Be Rejected

If \( H_0 \) is rejected, this means there is sufficient evidence to support the claim that the new bonus plan increases the sales volume. Therefore, we conclude that the mean sales volume under the new plan is greater than 14 automobiles per month, implying the plan's effectiveness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
When conducting hypothesis testing, the starting point is the null hypothesis. This hypothesis, denoted as \(H_0\), represents the default or original state, suggesting there is no change, no effect, or no difference. In the context of the automobile dealership problem, the null hypothesis posits that the new bonus plan does not influence the average sales volume.
For this scenario, the null hypothesis is mathematically expressed as \(H_0: \mu = 14\). Here, \(\mu\) represents the mean monthly sales of automobiles with the bonus plan.
The null hypothesis serves as a baseline, and it remains accepted unless there is sufficient evidence to suggest a significant change. Therefore, it is crucial to assess whether the observed results are purely by chance or indicate a genuine effect, which brings us to the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis, noted as \(H_a\), stands in opposition to the null hypothesis. It suggests that there is a detectable effect or a difference. In our dealership example, it states that the new bonus plan leads to an increase in the sales volume.
Formulated as \(H_a: \mu > 14\), the alternative hypothesis asserts that the mean sales are greater than the existing average of 14 automobiles per month.
The role of the alternative hypothesis is to express what the researcher aims to prove. If sufficient evidence is found, the null hypothesis is rejected in favor of the alternative hypothesis. This process involves statistical testing to gauge whether observed data supports the alternative hypothesis significantly.
Statistical Conclusion
After formulating both hypotheses, the next step is to draw a statistical conclusion based on the data collected. This decision hinges on whether you reject or fail to reject the null hypothesis.
If the null hypothesis \(H_0\) cannot be rejected, it suggests the data lacks evidence to show the new bonus plan increases sales. In simpler terms, we conclude the sales remain around 14 cars per month.
  • The hypothesis testing does not prove \(H_0\) true but merely indicates insufficient evidence against it.

Conversely, if \(H_0\) is rejected, it implies we have enough statistical evidence to support the alternative hypothesis. The conclusion here is that the mean sales exceed 14 cars, hence affirming the bonus plan's effectiveness.
  • Rejecting \(H_0\) gives confidence in adopting the new bonus strategy as it seemingly boosts sales volume.

Overall, the statistical conclusion helps in making informed decisions based on empirical evidence, guiding subsequent actions such as adopting or revising business strategies.

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Most popular questions from this chapter

Wall Street securities firms paid out record year-end bonuses of \(\$ 125,500\) per employee for 2005 (Fortune, February 6,2006 ). Suppose we would like to take a sample of employees at the Jones \(\&\) Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of \(\$ 125,500\) for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones \& Ryan were different from the population mean. b. Suppose a sample of 40 Jones \(\&\) Ryan employees showed a sample mean year- end bonus of \(\$ 118,000\), Assume a population standard deviation of \(\sigma=\$ 30,000\) and compute the \(p\) -value. c. With \(a=.05\) as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

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The chamber of commerce of a Florida Gulf Coast community advertises that area residential property is available at a mean cost of \(\$ 125,000\) or less per lot. Suppose a sample of 32 properties provided a sample mean of \(\$ 130,000\) per lot and a sample standard deviation of \(\$ 12,500 .\) Use a .05 level of significance to test the validity of the advertising claim.

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. A sample of 120 shoppers showed a sample mean waiting time of 8.4 minutes. Assume a population standard deviation of \(\sigma=3.2\) minutes. What is the \(p\) -value? c. \(\quad\) At \(a=.05,\) what is your conclusion? d. Compute a \(95 \%\) confidence interval for the population mean. Does it support your conclusion?

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