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The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in July 2016 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llll}4.44 & 3.15 & 2.42 & 3.96\end{array}\) \(\begin{array}{llll}4.51 & 4.17 & 3.69 & 4.62\end{array}\) \(\begin{array}{lll}3.80 & 3.36 & 3.85\end{array}\) The mean price of a Big Mac in the U.S. in July 2016 was \$5.04. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12.)

Short Answer

Expert verified
In summary, based on a hypothesis test with a significance level of \(\alpha=0.05\), we found convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price of $5.04. Our t-test with sample mean \(3.64\) and sample standard deviation \(0.74\) resulted in a test statistic of \(t = -6.45\), which is more extreme than the critical value of \(-1.812\). Hence, we reject the null hypothesis in favor of the alternative hypothesis.

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, we need to calculate the sample mean and sample standard deviation for the European Big Mac prices. For this, we can use the formulas: Sample mean (\(\bar{x}\)): \(\bar{x} = \frac{\sum x}{n}\) Sample standard deviation (s): \(s = \sqrt{\frac{\sum(x - \bar{x})^2}{n-1}}\) Using all the given European Big Mac prices, we have: \(\bar{x} = \frac{4.44 + 3.15 + 2.42 + 3.96 + 4.51 + 4.17 + 3.69 + 4.62 + 3.80 + 3.36 + 3.85}{11} = 3.64 \text{ (rounded to 2 decimal places)}\) \(s = \sqrt{\frac{(4.44 - 3.64)^2 + (3.15 - 3.64)^2 + \cdots + (3.85 - 3.64)^2}{11 - 1}} = 0.74 \text{ (rounded to 2 decimal places)}\)
02

Set Up Null and Alternative Hypotheses

Since we are testing whether the mean price of a Big Mac in Europe is less than the reported U.S. mean price ($5.04), the null and alternative hypotheses are as follows: Null Hypothesis (\(H_0\)): \(\mu = 5.04\) Alternative Hypothesis (\(H_a\)): \(\mu < 5.04\)
03

Calculate the Test Statistic

In this case the sample size is small (\(n = 11\)) and the population standard deviation is unknown, so we should use a t-test for this hypothesis test. The test statistic for a t-test can be calculated using the following formula: \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\) Using the sample mean (\(\bar{x} = 3.64\)), sample standard deviation (\(s = 0.74\)), and the U.S. mean price (\(\mu = 5.04\)), we have: \(t = \frac{3.64 - 5.04}{\frac{0.74}{\sqrt{11}}} = -6.45 \text{ (rounded to 2 decimal places)}\)
04

Find the Critical Value and Make a Conclusion

We will compare the test statistic (-6.45) to the critical value from the t-distribution table using the significance level (\(\alpha = 0.05\)) and degrees of freedom (\(n - 1 = 10\)). The corresponding critical value for a one-tailed t-test is -1.812. Since our test statistic is more extreme than the critical value (-6.45 < -1.812), we reject the null hypothesis in favor of the alternative hypothesis. Hence, there is convincing evidence to suggest that the mean price of a Big Mac in Europe in July 2016 is less than the reported U.S. price.

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Key Concepts

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

Sample Mean
When dealing with statistical data, the sample mean is an essential concept. It simply refers to the average of the values within a sample set. For instance, in the exercise concerning Big Mac prices in Europe, the sample mean has been calculated by adding all the prices and dividing by the number of observations. For the European Big Mac prices, you add up all given prices and then divide by 11—the total number of samples: \[ \bar{x} = \frac{4.44 + 3.15 + 2.42 + 3.96 + 4.51 + 4.17 + 3.69 + 4.62 + 3.80 + 3.36 + 3.85}{11} = 3.64 \text{ (rounded to 2 decimal places)} \] Hence, the sample mean, or \( \bar{x} \), is the core foundation to argue about average tendencies within the sample.
T-test
The t-test is a vital statistical test used when you want to compare the mean from your sample data against a known value, usually the population mean. In scenarios where our sample size is relatively small (often below 30), like in our European Big Mac exercise with just 11 observations, this is the preferred test. The critical step is utilizing the calculated sample mean and standard deviation to ascertain how different our sample mean is from the hypothesized population mean.The t-test formula is: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] Here, \( \mu \) is the population mean, \( \bar{x} \) is the sample mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. In our case, this test helps determine if the mean price of the Big Mac is statistically less than in the U.S.
Null and Alternative Hypotheses
Before any hypothesis test like the t-test can be conducted, you need to establish your hypotheses. These act as the statements that your test will evaluate.
  • Null Hypothesis \((H_0)\): This typically represents the status quo or no effect scenario. For our exercise, \(H_0\) is that the mean price of a Big Mac in Europe equals the U.S. mean price, or \( \mu = 5.04 \).
  • Alternative Hypothesis \((H_a)\): This represents the hypothesis you're testing against. Here, \(H_a\) is that the mean price of a Big Mac in Europe is less than the U.S. reported price, expressed as \( \mu < 5.04 \).
These hypotheses guide your analysis and interpretation.
Test Statistic
The test statistic is a valuable tool in hypothesis testing, as it standardizes our sample observation given the hypothesis being tested. Specifically, it measures how far our sample mean deviates from the population mean, given the variability in our sample.Calculating the test statistic for the t-test involves substituting the sample mean, the hypothesized population mean, and the sample standard deviation into the formula:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]In our exercise with European Big Mac prices, we have:\[ t = \frac{3.64 - 5.04}{\frac{0.74}{\sqrt{11}}} = -6.45 \text{ (rounded to 2 decimal places)} \]This number helps us understand how extreme our sample result is compared to what we'd expect if the null hypothesis were true.
Critical Value
In hypothesis testing, the critical value is the threshold that the test statistic is compared against to decide whether to reject the null hypothesis. This value comes from a statistical distribution—in our case, the t-distribution—based on the chosen significance level \( \alpha \).For a one-tailed test like our Big Mac price analysis, where \( \alpha = 0.05 \), and with degrees of freedom calculated as \( n - 1 = 10 \), the critical value from the t-distribution table is -1.812. Our test statistic (-6.45) is more negative than this critical value, meaning the test statistic falls within the rejection region of the t-distribution.Thus, in this instance, we reject the null hypothesis, supporting the alternative that Europe's average Big Mac price is less than the U.S. mean price.

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Most popular questions from this chapter

The paper "Alcohol Consumption, Sleep, and Academic Performance Among College Students" (Journal of Studies on Alcohol and Drugs [2009]: 355-363) describes a study of \(n=236\) students who were randomly selected from a list of students enrolled at a liberal arts college in the northeastern region of the United States. Each student in the sample responded to a number of questions about their sleep patterns. For these 236 students, the sample mean time spent sleeping per night was reported to be 7.71 hours and the sample standard deviation of the sleeping times was 1.03 hours. Suppose that you are interested in learning about the value of \(\mu,\) the population mean time spent sleeping per night for students at this college. The following table is similar to the table that appears in Example 12.4 . The "what you know" information has been provided. Complete the table by filling in the "how you know it" column.

The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).

USA TODAY (October 14,2016 ) reported that Americans spend 4.1 hours per weekday checking work e-mail. This was an estimate based on a survey of 1004 white-collar workers in the United States. a. Suppose that you would like to know if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. What would you need to assume about the sample in order to use the given sample data to answer this question? b. Given that any concerns about the sample were satisfactorily addressed, carry out a test to decide if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. Suppose that the sample standard deviation was \(s=1.3\) hours.

The paper "The Effects of Adolescent Volunteer Activities on the Perception of Local Society and Community Spirit Mediated by Self-Conception" (Advanced Science and Technology Letters [2016]: 19-23) describes a survey of a large representative sample of middle school children in South Korea. One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours. a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal. b. The sample size was not given in the paper, but the sample size was described as "large." Suppose that the sample size was 500 . Explain why it is reasonable to use a one-sample \(t\) confidence interval to estimate the population mean even though the population distribution is not approximately normal. c. Calculate and interpret a confidence interval for the mean number of hours spent in volunteer activities per year for South Korean middle school children. (Hint: See Example 12.7.)

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(\mathrm{df}=9, t=0.73\) b. Upper-tailed test, \(\mathrm{df}=10, t=0.5\) c. Lower-tailed test, \(n=20, t=-2.1\) d. Two-tailed test, \(n=40, t=1.7\)

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