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Chapter 9: Problem 109

Customers often complain about long waiting times at restaurants before the food is served. A restaurant claims that it serves food to its customers, on average, within 15 minutes after the order is placed. A local newspaper journalist wanted to check if the restaurant's claim is true. A sample of 36 customers showed that the mean time taken to serve food to them was \(15.75\) minutes with a standard deviation of \(2.4\) minutes. Using the sample mean, the journalist says that the restaurant's claim is false. Do you think the journalist's conclusion is fair to the restaurant? Use the \(1 \%\) significance level to answer this question.

Short Answer

Expert verified
Given the data, the journalist's conclusion is not fair to the restaurant, as there is no statistical evidence (at the 1% significance level) to refute the restaurant's claim that they serve food, on average, within 15 minutes.

Step by step solution

01

State the hypothesis

The null hypothesis \((H_0)\) is: the restaurant, on average, serves food within 15 minutes. Therefore, the mean serving time, \(\mu\), is present as \(H_0: \mu = 15\). The alternative hypothesis \((H_1)\) is: the restaurant does not serve food within 15 minutes. So, \(H_1: \mu \neq 15\). Here \(\mu\) represents the population mean not the sample mean.
02

Compute the test statistic

We need to calculate the test statistic using the sample mean \(15.75\), the population mean \(15\), the sample standard deviation \(2.4\), and the sample size \(36\). The formula for the test statistic \(z\) in a Z-test is \[z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})}\] Substituting the numbers into the formula, we get \[ z = \frac{(15.75 - 15)}{(2.4 / \sqrt{36})} \] which evaluates to \(z = 1.875\).
03

Determine the p-value

The p-value is the probability of observing a sample statistic as extreme as the test statistic. Since the hypothesis test is two-tailed (the alternative hypothesis is \(\mu \neq 15\)), the p-value is the probability that a z-score is less than -1.875 or greater than 1.875. Referring to standard statistical Z-tables, the two-tailed p-value corresponds to the z-score of 1.875 is about 0.061.
04

Compare the p-value to the significance level

The significance level is given as 1% (or 0.01). Since the p-value (0.061) is greater than the significance level (0.01), we fail to reject the null hypothesis.
05

Draw Conclusion

Since the null hypothesis was not rejected, it means that there is not enough evidence to claim that the restaurant does not serve food, on average, within 15 minutes. Thus, saying that the restaurant's claim is false would indeed be unfair based on this test.

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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 \(H_0\), represents a default or initial assumption that there is no effect or difference. In our context, it is the statement that the restaurant serves food within 15 minutes, on average. We express this mathematically as \(H_0: \mu = 15\), where \(\mu\) is the population mean of the serving time. The null hypothesis sets a baseline or standard that the data is tested against to determine if there's significant evidence to indicate a deviation.
Alternative Hypothesis
The alternative hypothesis, symbolized by \(H_1\), challenges the null hypothesis by proposing a different reality. It suggests the presence of an effect or difference. For the restaurant scenario, the alternative hypothesis is that the restaurant does not serve food within 15 minutes, on average. Mathematically, this is noted as \(H_1: \mu eq 15\). This hypothesis is crucial for testing as it opens the possibility to discover new insights or changes in data trends beyond the null hypothesis鈥檚 perspective.
P-value
The p-value is a fundamental concept in hypothesis testing. It measures the strength of the evidence against the null hypothesis. Specifically, it's the probability of observing a test statistic as extreme as鈥攐r more extreme than鈥攖he one computed from the sample data, assuming the null hypothesis is true. In this example, a calculated z-score of 1.875 corresponds to a p-value of approximately 0.061. A smaller p-value indicates stronger evidence against the null hypothesis, whereas a larger p-value indicates weaker evidence.
Significance Level
The significance level, often represented as \(\alpha\), is a threshold set by the researcher to decide when to reject the null hypothesis. Commonly used significance levels are 0.05, 0.01, etc. They reflect the probability of incorrectly rejecting the null hypothesis, known as a Type I error. Here, a 1% or 0.01 significance level is used. This means that there is a 1% risk of concluding that the restaurant does not serve food within 15 minutes when it actually does. If the p-value is less than \(\alpha\), the null hypothesis is rejected.
Z-test
A Z-test is a statistical test used to determine whether two population means are different when the variances are known and the sample size is large. It helps assess whether there is enough evidence to reject the null hypothesis. The formula for the test statistic \(z\) is:\[z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})}\]where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size. For the restaurant, \(z = 1.875\), meaning the sample mean is 1.875 standard deviations above the claimed mean.
Two-tailed Test
In hypothesis testing, a two-tailed test checks for deviations in two directions, either higher or lower than the expected outcome. It is used when the alternative hypothesis includes values both greater and less than the mean specified in the null hypothesis. For our exercise, since \(H_1: \mu eq 15\), values greater and less than 15 are of interest. This approach makes the test rigorous as it considers both extremes, increasing the comprehensiveness of the inference drawn from the data. This testing method directly influenced the calculation of the p-value as it considers the probability of extremes on both tails of the normal distribution.

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Most popular questions from this chapter

In an observational study at Turner Field in Atlanta, Georgia, \(43 \%\) of the men were observed not washing their hands after going to the bathroom (see Exercise \(7.80\) ). Suppose that in a random sample of 95 men who used the bathroom at Camden Yards in Baltimore, Maryland, 26 did not wash their hands. a. Using the critical-value approach and \(\alpha=.10\), test whether the percentage of all men at Camden Yards who use the bathroom and do not wash their hands is less than \(43 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.10 ?\)

Consider the following null and alternative hypotheses: $$ H_{0}: p=.44 \text { versus } H_{1}: p<.44 $$ A random sample of 450 observations taken from this population produced a sample proportion of \(.39 .\) a. If this test is made at the \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01\) ? What if \(\alpha=.025\) ?

An carlier study claimed that U.S. adults spent an average of 114 minutes with their families per day. A recently taken sample of 25 adults from a city showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the times spent by adults with their families have an approximately normal distribution. a. Using the \(1 \%\) significance level, test whether the mean time spent currently by all adults with their families in this city is different from 114 minutes a day. b. Suppose the probability of making a Type I error is zero. Can you make a decision for the test of part a without going through the five steps of hypothesis testing? If yes, what is your decision? Explain.

Since 1984 , all automobiles have been manufactured with a middle tail-light. You have been hired to answer the question. Is the middle tail-light effective in reducing the number of rear-end collisions? You have available to you any information you could possibly want about all rear-end collisions involving cars built before 1984 . How would you conduct an experiment to answer the question? In your answer, include things like (a) the precise meaning of the unknown parameter you are testing: (b) \(H_{0}\) and \(H_{1} ;\) (c) a detailed explanation of what sample data you would collect to draw a conclusion; and (d) any assumptions you would make, particularly about the characteristics of cars built before 1984 versus those built since 1984 .

In Las Vegas and Atlantic City, New Jersey, tests are performed often on the various gaming devices used in casinos. For example, dice are often tested to determine if they are balanced. Suppose you are assigned the task of testing a die, using a two-tailed test to make sure that the probability of a 2 -spot is \(1 / 6 .\) Using the \(5 \%\) significance level, determine how many 2 -spots you would have to obtain to reject the null hypothesis when your sample size is \(\begin{array}{lll}\text { a. } 120 & \text { b. } 1200 & \text { c. } 12,000\end{array}\) Calculate the value of \(\hat{p}\) for each of these three cases. What can you say about the relationship between (1) the difference between \(\hat{p}\) and \(1 / 6\) that is necessary to reject the null hypothesis and (2) the sample size as it gets larger?
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