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Chapter 9: Problem 2

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 research 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

Expert verified
a. \( H_0: \mu = 14 \), \( H_a: \mu > 14 \). b. The bonus plan may not affect sales volume significantly. c. The bonus plan likely increases sales volume significantly.

Step by step solution

01

Identify the Null Hypothesis

The null hypothesis, denoted as \( H_0 \), represents the current condition or status quo that there is no effect or no change due to the new bonus plan. In this context, it refers to the mean sales volume staying the same. Therefore, our null hypothesis is \( H_0: \mu = 14 \), where \( \mu \) is the mean sales volume under the new bonus plan.
02

Formulate the Alternative Hypothesis

The alternative hypothesis, denoted by \( H_a \), indicates the presence of an effect or a difference. In this case, it suggests that the new bonus plan increases sales volume. Thus, the alternative hypothesis is \( H_a: \mu > 14 \).
03

Conclusion When Null Hypothesis Cannot Be Rejected

When the null hypothesis \( H_0 \) cannot be rejected, it means there is not enough statistical evidence to support an increase in sales volume due to the new bonus plan. The manager may conclude that the bonus plan does not have a significant effect, and any observed changes might be due to random chance.
04

Conclusion When Null Hypothesis Can Be Rejected

If the null hypothesis \( H_0 \) can be rejected, it suggests that there is significant statistical evidence that the new bonus plan increases the mean sales volume. The manager can infer that the plan is effective at increasing sales.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis (\(H_0\)) serves as a statement of no effect or no change. It is the baseline you test against, and it's assumed true until you can provide evidence otherwise.
In our scenario, the null hypothesis (\(H_0\)) claims that the mean sales volume remains unchanged at 14 automobiles per month. This hypothesis represents the dealership manager's starting point; that the new bonus plan does not affect sales volume. By assuming everything stays constant, the null hypothesis allows us to test whether any observed change is statistically significant or merely due to randomness.
Remember:
  • The null hypothesis is about proving the status quo.
  • It's typically formatted as an equation: \(H_0: \mu = 14\), indicating that the mean sales volume doesn't increase.
Alternative Hypothesis
When you suspect a change or effect, you consider the alternative hypothesis (\(H_a\)). It reflects a potential outcome contrary to the null hypothesis. In our context, the alternative hypothesis proposes that the new bonus plan has increased the sales volume.
The purpose of \(H_a\) is to suggest that something different has happened, such as the mean sales volume surpassing 14 automobiles per month with the new plan applied. This hypothesis is the focal point of the investigation, as it tests whether the plan has a measurable impact on sales.
Key points include:
  • It opposes the null hypothesis.
  • In our example, \(H_a: \mu > 14\), which means the new bonus plan increases average sales.
  • It represents what you're trying to prove.
Statistical Evidence
Collecting enough statistical evidence is crucial for deciding between the null and alternative hypotheses. Statistical evidence refers to data from your sample size that shows whether your hypothesis can be supported or not.
In this exercise, statistical evidence would derive from the sales figures recorded under the new bonus plan. A significant change in these figures compared to when no bonus plan was implemented helps infer the plan's effectiveness.
Here鈥檚 what to consider:
  • Evidence is crucial to either reject \(H_0\) or support \(H_a\).
  • The strength of evidence depends on sample size and variability of results.
  • If evidence is strong enough against \(H_0\), it might lead to its rejection, suggesting that the bonus plan increases sales.
Ultimately, solid statistical evidence guides you to an informed decision about whether the new strategy influences sales outcomes.

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Most popular questions from this chapter

Nielsen reported that young men in the United States watch 56.2 minutes of prime-time TV daily (The Wall Street Journal Europe, November 18,2003 ). A researcher believes that young men in Germany spend more time watching prime- time TV. A sample of German young men will be selected by the researcher and the time they spend watching TV in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. \\[ \begin{array}{l} H_{0}: \mu \leq 56.2 \\ H_{\mathrm{a}}: \mu>56.2 \end{array} \\] a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=15 \\ H_{\mathrm{a}}: \mu \neq 15 \end{array} \\] A sample of 50 provided a sample mean of \(14.15 .\) The population standard deviation is \(3 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United States: 2006 ). Being from the Midwest, you believe milk consumption is higher there and wish to support your opinion. A sample of 16 individuals from the midwestern town of Webster City showed a sample mean annual consumption of 24.1 gallons with a standard deviation of \(s=4.8\) a. Develop a hypothesis test that can be used to determine whether the mean annual consumption in Webster City is higher than the national mean. b. What is a point estimate of the difference between mean annual consumption in Webster City and the national mean? c. \(\quad\) At \(\alpha=.05,\) test for a significant difference. What is your conclusion?

The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is \(\$ 600\) or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of weekend guest bills to test the manager's claim. a. Which form of the hypotheses should be used to test the manager's claim? Explain. \\[ \begin{array}{lll} H_{0}: \mu \geq 600 & H_{0}: \mu \leq 600 & H_{0}: \mu=600 \\ H_{\mathrm{a}}: \mu<600 & H_{\mathrm{a}}: \mu>600 & H_{\mathrm{a}}: \mu \neq 600 \end{array} \\] b. What conclusion is appropriate when \(H_{0}\) cannot be rejected? c. What conclusion is appropriate when \(H_{0}\) can be rejected?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=100 \\ H_{\mathrm{a}}: \mu \neq 100 \end{array} \\] A sample of 65 is used. Identify the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.05\) a. \(\bar{x}=103\) and \(s=11.5\) b. \(\bar{x}=96.5\) and \(s=11.0\) c. \(\bar{x}=102\) and \(s=10.5\)
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