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The paper "Sodium content of Lunchtime Fast Food Purchases at Major U.S. Chains" (Archives of Internal Medicine [2010]: \(732-734\) ) reported that for a random sample of 850 meal purchases made at Burger King, the mean sodium content was \(1685 \mathrm{mg},\) and the standard deviation was \(828 \mathrm{mg}\). For a random sample of 2107 meal purchases made at Mel)onald's, the mean sodium content was \(1477 \mathrm{mg},\) and the standard deviation was \(812 \mathrm{mg}\). Based on these data, is it reasonable to conclude that there is a difference in mean sodium content for meal purchases at Burger King and meal purchases at McDonald's? Use \(\alpha=0.05\).

Short Answer

Expert verified
In conclusion, based on the two-sample t-test results, there is a significant difference in mean sodium content for meal purchases at Burger King and McDonald's at a significance level of \(\alpha = 0.05\).

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that there is no significant difference between the true mean sodium content at Burger King (\(\mu_{BK}\)) and at McDonald's (\(\mu_{M}\)). The alternative hypothesis (\(H_a\)) states that there is a significant difference. $$ H_0: \mu_{BK} = \mu_{M} $$ $$ H_a: \mu_{BK} \ne \mu_{M} $$
02

Determine the critical t-value

Given the information in the problem, we perform a two-sample t-test, and we have a significance level (\(\alpha\)) of 0.05. Since this is a two-tailed test, we need to find the t-value at \(\alpha/2\) for each tail. Degrees of freedom (\(df\)) can be determined by: $$ df = n_{1} + n_{2} - 2 = 850 + 2107 - 2 = 2955 $$ Using a t-table or a calculator, we find the critical t-value to be approximately \(\pm 1.96\).
03

Calculate the test statistic

We will use the pooled t-test formula: $$ t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_1^2}{n_{1}} + \frac{s_2^2}{n_{2}}}} $$ Where \(\bar{x}_1, \bar{x}_2\) are the respective sample means, \(s_1, s_2\) are the respective sample standard deviations, and \(n_1, n_2\) are the respective sample sizes. Plugging in the given values: $$ t = \frac{1685 - 1477}{\sqrt{\frac{828^2}{850} + \frac{812^2}{2107}}} \approx 10.39 $$
04

Compare the test statistic to the critical t-value

We have calculated the test statistic \(t \approx 10.39\). Now we will compare it to the critical t-value \(\pm 1.96\). Since \(10.39 > 1.96\), we reject the null hypothesis.
05

Draw the conclusion

Based on the results of the two-sample t-test, we reject the null hypothesis that there is no difference in mean sodium content for meal purchases at Burger King and McDonald's. Therefore, it is reasonable to conclude that there is a significant difference in mean sodium content between these two fast-food chains, at a significance level of \(\alpha = 0.05\).

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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 symbolized as \(H_0\), is a critical concept in statistical testing that represents the assumption that no effect or no difference exists in a particular situation. For example, when comparing the mean sodium content of meals between Burger King and McDonald's, our null hypothesis is that there is no significant difference between their true mean sodium levels. Statistically, we denote this as \(H_0: \mu_{BK} = \mu_{M}\). This hypothesis serves as a starting point for our analysis, and the objective is to challenge it with sample data to see if there's enough evidence to support the alternative hypothesis.

Understanding the null hypothesis is essential because it establishes a baseline that enables us to measure the probability of observing our sample data, assuming the null hypothesis is true. If the probability is low enough, we have a reason to reject the null and accept that an effect or difference likely exists.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis, denoted as \(H_a\) or \(H_1\), reflects the presence of an effect or difference that the researcher is attempting to detect. For our two-sample t-test comparing the sodium content of meals at Burger King and McDonald's, the alternative hypothesis is that there is a significant difference in mean sodium content, which we write as \(H_a: \mu_{BK} eq \mu_{M}\).

The alternative hypothesis is what we believe might truly describe or explain our data. This hypothesis is considered if there is significant statistical evidence to reject the null hypothesis. Researchers must clearly define both null and alternative hypotheses before conducting the test to avoid bias in their analysis and ensure that they are testing what they intend to.
Test Statistic
The test statistic in a two-sample t-test provides a numerical value that helps in making decisions about the hypotheses. It measures how far our sample statistic is from the null hypothesis value in terms of standard errors. If the test statistic falls into a critical region, which is unlikely if the null hypothesis is true, we then have grounds to reject the null.

For our example involving fast-food chains, we calculate the test statistic using the formula \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_1^2}{n_{1}} + \frac{s_2^2}{n_{2}}}}\) where the sample means \(\bar{x}_{1}, \bar{x}_{2}\), the sample standard deviations \(s_1, s_2\), and the sample sizes \(n_1, n_2\) are known. A test statistic that falls outside of the range defined by the critical t-values suggests that the observed sample differences are statistically significant.
Pooled Standard Deviation
Pooled standard deviation is an estimate of the common standard deviation of two groups that is used when conducting a two-sample t-test under the assumption that the population variances are equal. It combines the variances of each group, weighted by their respective degrees of freedom. This pooled value is then used to standardize the difference between the two sample means, which allows us to calculate the test statistic.

In scenarios where the assumption of equal variances is not valid, a different approach called 'Welch’s t-test' is employed which does not assume equal population variances. The formula for pooled standard deviation is not used in the test statistic for the sample problem provided since the calculated test statistic uses the sample standard deviations directly.
Degrees of Freedom
Degrees of freedom (df) often confuse students, but this concept is essential in the context of hypothesis testing. It refers to the number of independent values that have the freedom to vary when computing a statistic, such as a mean or variance. In a two-sample t-test, the degrees of freedom are calculated based on the sample sizes of the two groups. The formula for degrees of freedom is \(df = n_{1} + n_{2} - 2\).

Knowing the degrees of freedom is crucial because it allows us to select the appropriate values from the t-distribution when determining our critical t-values and when interpreting the test statistic. In our exercise example, with sample sizes of 850 and 2107, we have 2955 degrees of freedom. These degrees of freedom are then used along with the significance level to find the critical t-value, which serves as a threshold for deciding whether to reject the null hypothesis.

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Most popular questions from this chapter

Herc's one to sink your teeth into: The authors of the article "Analysis of Food Crushing Sounds During Mastication: Total Sound Level Studies" (Journal of Texture Studies [1990]: \(165-178\) ) studicd the nature of sounds generated during eating. Peak loudness was measured (in decibels at \(20 \mathrm{~cm}\) away \()\) for both open-mouth and closedmouth chewing of potato chips and of tortilla chips. A sample of size 10 was used for each of the four possible combinations (such as closed-mouth potato chip, and so on). We are not making this up! Summary values taken from plots given in the article appear in the accompanying table. For purposes of this exercise, suppose that it is reasonable to regard the peak loudness distributions as approximately normal. a. Construct a 95\% confidence interval for the difference in mean peak loudness between open-mouth and closedmouth chewing of potato chips. Be sure to interpret the resulting interval. b. For closed-mouth chewing (the recommended method!), construct a \(95 \%\) confidence interval for the difference in mean peak loudness between potato chips and tortilla chips.

The paper "Short-Term Sleep Loss Decreases Physical Activity Under Free-Living Conditions but Does Not Increase Food Intake Under Time-Deprived Laboratory Conditions in Healthy Men" (American Journal of Clinical Nutrition [2009]: \(1476-1483\) ) describes an experiment in which 30 male volunteers were assigned at random to one of two slecp conditions. Men in the 4 -hour group slept 4 hours per night for two nights. Men in the 8 -hour group slept 8 hours per night for two nights. On the day following these two nights, the men recorded food intake. The researchers reported that there was no significant difference in mean calorie intake for the two groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means.

Do female college students spend more time watching TV than male college students? This was one of the questions investigated by the authors of the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Health Education Journal [2010]: 116-125). Each student in a random sample of 46 male students at a university in England and each student in a random sample of 38 female students from the same university kept a diary of how he or she spent time over a 3 -week period. For the sample of males, the mean time spent watching TV per day was 68.2 minutes, and the standard deviation was 67.5 minutes. For the sample of females, the mean time spent watching TV per day was 93.5 minutes, and the standard deviation was 89.1 minutes. Is there convincing evidence that the mean time female students at this university spend watching TV is greater than the mean time for male students? Test the appropriate hypotheses using \(\alpha=0.05\).

The article "An Alternative Vote: Applying Science to the Teaching of Science" (The Economist, May 12,2011 ) describes an experiment conducted at the University of British Columbia. A total of 850 engineering students enrolled in a physics course participated in the experiment. Students were randomly assigned to one of two experimental groups. Both groups attended the same lectures for the first 11 weeks of the semester. In the twelfth week, one of the groups was switched to a style of teaching where students were expected to do reading assignments prior to class, and then class time was used to focus on problem solving, discussion, and group work. The second group continued with the traditional lecture approach. At the end of the twelfth week, students were given a test over the course material from that week. The mean test score for students in the new teaching method group was 74 , and the mean test score for students in the traditional lecture group was 41 . Suppose that the two groups each consisted of 425 students. Also suppose that the standard deviations of test scores for the new teaching method group and the traditional lecture method group were 20 and \(24,\) respectively. Estimate the difference in mean test score for the two teaching methods using a \(95 \%\) confidence interval. Be sure to give an interpretation of the interval.

The paper "Effects of Caffeine on Repeated Sprint Ability, Reactive Agility Time, Sleep and Next Day Performance" (Journal of Sports Medicine and Physical Fitness [2010]: \(455-464)\) describes an experiment in which male athlete volunteers who were considered low caffeine consumers were assigned at random to one of two cxpcrimental groups. Those assigned to the caffeine group drank a beverage which contained caffeine one hour before an excrcise session. Those in the no-caffeine group drank a beverage that did not contain caffeine. During the exercise session, each participant performed a test that measured reactive agility. The researchers reported that there was \(n 0\) significant difference in mean reactive agility for the two experimental groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means.

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