91Ó°ÊÓ

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. $$ \begin{array}{rrrrrr} \text { Variable } & \text { M Mean StDev } & \text { SEMean } & \text { ZP- Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40-2.14 & 0.016 \end{array} $$ What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of \(.01\) ? What significance level and conclusion would you recommend?

Step by step solution

01

Define the Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis (H_0) is that the average lifetime of the bulbs is 750 hours. The alternative hypothesis (H_1) is that the average lifetime is less than 750 hours. Mathematically, these are expressed as follows:- H_0: \mu = 750- H_1: \mu < 750where \(\mu\) is the true mean lifetime of the bulbs.

02

Identify the Test Statistic and Significance Level

The test statistic is a Z-score, and from the MINITAB output, we see the calculated Z-value is -2.14. We'll be considering two significance levels: \(\alpha = 0.05\) and \(\alpha = 0.01\).

03

Find Critical Values for Given Significance Levels

Using the standard normal distribution table, find the critical Z-value for a significance level of 0.05. The critical value for \(\alpha = 0.05\) in a one-tailed test is approximately -1.645. For \(\alpha = 0.01\), the critical value is approximately -2.33.

04

Compare Test Statistic with Critical Values

Now, we compare the Z-score from our test to the critical values:- At \(\alpha = 0.05\), compare -2.14 with -1.645. Since -2.14 is less than -1.645, we reject the null hypothesis at the 0.05 significance level.- At \(\alpha = 0.01\), compare -2.14 with -2.33. Since -2.14 is greater than -2.33, we fail to reject the null hypothesis at the 0.01 significance level.

05

Determine P-value Interpretation

From the output, the p-value is 0.016. Once again, compare it to the significance levels:- At \(\alpha = 0.05\), since 0.016 < 0.05, we reject the null hypothesis.- At \(\alpha = 0.01\), since 0.016 > 0.01, we fail to reject the null hypothesis.

06

Conclusion and Recommendation

At a significance level of 0.05, the evidence suggests the true mean lifetime is less than 750 hours, and we reject the null hypothesis. At 0.01, the evidence is not strong enough to reject the null. A significance level of 0.05 is often standard in many fields, so we recommend using it and concluding there is sufficient evidence to suggest the advertised lifetime is overstated.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Significance Level

In hypothesis testing, the significance level, often denoted as \( \alpha \), is a threshold set by the researcher that defines the maximum acceptable risk of making a Type I error. A Type I error occurs when we incorrectly reject a true null hypothesis. For example, in the context of the lightbulb scenario, if we choose a significance level of \( 0.05 \), we are saying there is a 5% risk of concluding that the average lifetime of the bulbs is less than 750 hours when it is not.

• A common significance level is \( \alpha = 0.05 \).
• A more stringent level is \( \alpha = 0.01 \), which reduces the risk of a Type I error but also requires stronger evidence to reject the null hypothesis.

Selecting the right significance level depends on the context and the potential consequences of making an incorrect decision.

Null and Alternative Hypotheses

In hypothesis testing, the first step is to define the null and alternative hypotheses. These hypotheses are statements about a population parameter that you want to test.The **null hypothesis** (\( H_0 \)) is a statement of no effect or no difference—it is the status quo or the claim that is initially assumed to be true. For the lightbulb case, the null hypothesis is that the average lifetime of the bulbs is 750 hours: \( H_0: \mu = 750 \).The **alternative hypothesis** (\( H_1 \)) is what you want to prove. It is the statement that indicates the presence of an effect or difference. Here, the alternative hypothesis is that the average lifetime is less than the advertised 750 hours: \( H_1: \mu < 750 \).

• The null and alternative hypotheses are mutually exclusive.
• Considering the direction, this is a one-tailed test because we are only interested in deviations below 750 hours.

Z-test

A Z-test is a type of hypothesis test used for determining if there is a significant difference between a sample mean and the population mean when the population variance is known or the sample size is large (typically \( n \geq 30 \)). In the lightbulb example, the Z-test is performed since we have a fairly large sample size of 50 bulbs.The Z-score, or Z-value, is a measure of how many standard deviations an element is from the mean. It is calculated using the formula: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where:

• \( \bar{x} \) is the sample mean.
• \( \mu \) is the population mean.
• \( \sigma \) is the population standard deviation.
• \( n \) is the sample size.

In this case, the Z-value was found to be -2.14, indicating how far 738.44 hours (the sample mean) is from the 750 hours, in terms of standard deviations.

P-value Interpretation

The p-value in hypothesis testing helps us determine the strength of our results. It is the probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is true.For the lightbulb problem, the calculated p-value is 0.016. This value helps in deciding whether to reject or fail to reject the null hypothesis.

• If the p-value is less than or equal to the chosen significance level (e.g., 0.05), we reject the null hypothesis.
• If the p-value is greater than the significance level, we fail to reject the null hypothesis.

In this scenario:- At \( \alpha = 0.05\), since 0.016 < 0.05, there is enough evidence to reject the null hypothesis and conclude that the average lifetime is indeed less than 750 hours.- At \( \alpha = 0.01\), since 0.016 > 0.01, we do not have strong enough evidence to reject the null hypothesis at that significance level.The p-value is a powerful tool for interpreting the results of hypothesis tests and making informed decisions based on data.