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Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using Minitab, resulting in the accompanying output. \(\begin{array}{lrrrrrr}\text { Variable } & N & \text { Mean } & \text { StDev } & \text { SE Mean } & \text { Z } & \text { P-Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40 & -2.14 & 0.016\end{array}\) What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of .01? What significance level and conclusion would you recommend?

Short Answer

Expert verified
Reject the null at 0.05, not at 0.01; recommend 0.05 level.

Step by step solution

01

Identify Null and Alternative Hypotheses

The problem states that we need to determine if the true average lifetime is less than 750 hours. Thus, the hypotheses are: \[ H_0: \mu = 750 \text{ hours} \] \[ H_a: \mu < 750 \text{ hours} \] where \( \mu \) is the true average lifetime of the lightbulbs. This is a one-tailed test.
02

Analyze Minitab Output

The Minitab output provides the Z statistic and P-value. Here, the Z statistic is -2.14, and the P-value is 0.016. These values will help us determine whether to reject or not reject the null hypothesis.
03

Conclusion at Significance Level 0.05

At a significance level of 0.05, we reject the null hypothesis if the p-value is less than 0.05. The given p-value is 0.016, which is less than 0.05. Therefore, we reject the null hypothesis at the 0.05 significance level.
04

Conclusion at Significance Level 0.01

At a significance level of 0.01, we reject the null hypothesis if the p-value is less than 0.01. The given p-value is 0.016, which is greater than 0.01. Therefore, we do not reject the null hypothesis at the 0.01 significance level.
05

Recommended Significance Level and Conclusion

Given the importance of certainty in product claims and customer decisions, and since we can reject the null hypothesis at a 0.05 level but not at 0.01, it would be recommended to use a 0.05 significance level. At this level, we conclude that the average lifetime is significantly less than 750 hours.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a foundational element in hypothesis testing. It represents a statement of no effect or no difference and is what you test against your data. In the context of our lightbulb problem, the null hypothesis is: \( H_0: \mu = 750 \) hours. This hypothesis asserts that the true average lifetime of the lightbulbs is 750 hours, as advertised. Think of the null hypothesis as a default position we start with before testing any evidence. We assume that the advertisement about the lightbulb's lifespan is accurate unless data suggest otherwise. The goal of hypothesis testing is to determine whether there is enough statistical evidence to reject the null hypothesis in favor of an alternative. To make this decision, we rely on calculating a p-value or a test statistic, like the Z statistic provided in the exercise.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), stands in contrast to the null hypothesis. It is what you might believe to be true or hope to prove true. For the lightbulb scenario, the alternative hypothesis is: \( H_a: \mu < 750 \) hours.This hypothesis suggests that the true average lifetime of the lightbulbs is less than the advertised 750 hours. It is a one-sided hypothesis because it only considers the possibility of a decrease in the average lifetime and not an increase. Testing the alternative hypothesis involves using statistical evidence to challenge the null hypothesis. If the data strongly support the possibility that \( \mu \) is indeed less than 750 hours, we reject the null hypothesis. Otherwise, we cannot conclude the alternative hypothesis is true. This testing informs decisions like whether to go forward with a purchase.
Significance Level
The significance level, often denoted \( \alpha \), is a threshold that determines how strong the evidence must be before we reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. Common significance levels are 0.05, 0.01, and 0.10, which translate to a 5%, 1%, and 10% chance, respectively, of making a Type I error.In the lightbulb exercise, if we choose a significance level of 0.05, we're willing to accept a 5% chance of incorrectly concluding that the bulbs last less than 750 hours when they don't. A smaller significance level, like 0.01, requires stronger evidence to reject the null hypothesis, reducing the chance of error but may also make it harder to detect a true effect if one exists.Deciding on a significance level involves balancing the risk of errors with the need for evidence. If the consequence of making an error is significant, such as misleading a customer, a lower alpha might be preferred. In this case, using \( \alpha = 0.05 \) was recommended because the data provided sufficient evidence at this level, but not at \( \alpha = 0.01 \). This illustrates how the significance level guides hypothesis testing decisions.

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Most popular questions from this chapter

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Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. The article "Properties of Waste Silk Short Fiber/Cellulose Green Composite Films" ( \(J\). of Composite Materials, 2012: 123-127) reported that for a sample of 10 specimens with \(2 \%\) fiber content, the sample mean tensile strength (MPa) was \(51.3\) and the sample standard deviation was 1.2. Suppose the true average strength for \(0 \%\) fibers (pure cellulose) is known to be \(48 \mathrm{MPa}\). Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value?

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