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A student organization uses the proceeds from a soft drink vending machine to finance its activities. The price per can was \(\$ 0.75\) for a long time, and the mean daily revenue during that period was \(\$ 75.00\). The price was recently increased to \(\$ 1.00\) per can. A random sample of \(n=20\) days after the price increase yielded a sample mean daily revenue and sample standard deviation of \(\$ 70.00\) and \(\$ 4.20\), respectively. Does this information suggest that the mean daily revenue has decreased from its value before the price increase? Test the appropriate hypotheses using \(\alpha=\) \(0.05 .\)

Short Answer

Expert verified
In conclusion, the t-test results (t = 5.345, critical value = 1.729) indicate that the mean daily revenue has significantly decreased after the price increase for soft drinks. With a significance level of 伪 = 0.05, we reject the null hypothesis and conclude that the mean daily revenue has decreased. The student organization should consider adjusting the price again or exploring other ways to increase revenue.

Step by step solution

01

Define the hypotheses

We define the null hypothesis (H鈧) and alternative hypothesis (H鈧) as follows: H鈧: 渭鈧 - 渭鈧 = 0 (There is no difference in mean daily revenue before and after the price increase) H鈧: 渭鈧 - 渭鈧 > 0 (The mean daily revenue has decreased after the price increase) Where 渭鈧 is the mean daily revenue before the price increase and 渭鈧 is the mean daily revenue after the price increase.
02

Compute the test statistic

The t-test statistic is given by the formula: \[t = \frac{(\bar{x_1} - \bar{x_2}) - D鈧}{\frac{s}{\sqrt{n}}}\] Where: - \(\bar{x_1}\) is the mean daily revenue before the price increase (\$75.00) - \(\bar{x_2}\) is the mean daily revenue after the price increase (\$70.00) - \(D鈧\) is the hypothesized difference between the two means (0) - \(s\) is the sample standard deviation (4.20) - \(n\) is the sample size (20) Using the data provided: \[t = \frac{(75 - 70) - 0}{\frac{4.2}{\sqrt{20}}}\]
03

Calculate the t-test statistic

Now we will perform the calculation for the t-test statistic: \[t = \frac{5}{\frac{4.2}{\sqrt{20}}}\] \[t 鈮 5.345\]
04

Determine the critical value

We will use the t-distribution table to find the critical value for a one-tailed test with a significance level of 伪 = 0.05 and degrees of freedom (df) = n - 1 = 19. Looking up the value in the table, we find that the critical value is approximately 1.729.
05

Compare the test statistic with the critical value

We will now compare the t-test statistic (5.345) with the critical value (1.729): 5.345 > 1.729 Since the test statistic is greater than the critical value, we reject the null hypothesis (H鈧).
06

Conclusion

Based on the results of the t-test, we have enough evidence to conclude that the mean daily revenue has decreased after the price increase. The student organization should consider adjusting the price again or exploring other ways to increase revenue.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial in hypothesis testing. These hypotheses are the backbone of any statistical test and are used to ascertain if there is significant evidence to support a certain belief about a population parameter.

In our example, the null hypothesis (\(H_0\)) is that the mean daily revenue from the vending machine (\(\mu_1\-\mu_2\) did not change after the price increase. Conversely, the alternative hypothesis (\(H_1\) states that the mean daily revenue actually decreased after the price change. This can also be understood as the hypothesis that expects a change or difference, here implying that the mean daily revenue after the price hike (\(\mu_2\) is less than before the hike (\(\mu_1\) In hypothesis testing, we initially assume that the null hypothesis is true and then use statistical evidence to determine whether there is enough data to reject this assumption in favor of the alternative.
T-test Statistic Calculation
Calculating the t-test statistic involves using sample data to measure the difference between the observed sample mean and the proposed population mean, relative to the variability in the sample.

The t-test statistic is calculated using the formula: \[t = \frac{(\bar{x_1} - \bar{x_2}) - D_0}{\frac{s}{\sqrt{n}}}\] This formulation takes into account the sample means (\(\bar{x_1}\) and \(\bar{x_2}\)), the hypothesized difference in means (\(D_0\)), the sample standard deviation (\(s\)) and the sample size (\(n\)). In our exercise, the t-test statistic was used to determine whether the observed sample mean revenue after the price increase significantly differs from the previous mean revenue.
Significance Level (Alpha)
The significance level, denoted as alpha (\(\alpha\)), is a threshold used to determine the probability of committing a Type I error, which occurs when the null hypothesis is incorrectly rejected.

A commonly used \(\alpha\) value is 0.05, meaning there is a 5% chance of rejecting a true null hypothesis. This level of significance is a balance between being too lenient and too strict, allowing researchers to be reasonably confident in their decision to reject or fail to reject the null hypothesis. In the provided exercise, an \(\alpha\) of 0.05 was chosen to test the hypothesis about the average revenue from the vending machine.
Degrees of Freedom
Degrees of freedom, often abbreviated as 'df', is a concept closely tied with the estimation of population parameters. It refers to the number of values in a calculation that are free to vary.

In the context of the t-test, the degrees of freedom are calculated as the sample size (\(n\)) minus one (\(df = n - 1\)), which in our example would result in 19 degrees of freedom. This number is essential for determining the correct critical value from the t-distribution table and helps to account for the sample size in the estimation process. The degrees of freedom basically reflect the sample size's contribution to the precision of the estimated population parameter.
Critical Value Determination
The critical value is a point on the distribution of the test statistic beyond which we would reject the null hypothesis. It depends on the chosen significance level and the degrees of freedom.

To determine the critical value, you would usually refer to a t-distribution table or use statistical software. This value serves as a comparison point for our calculated t-test statistic: if the t-test statistic is greater than the critical value, the null hypothesis is rejected. In the example provided, with an \(\alpha\) of 0.05 and 19 degrees of freedom, the critical value was found to be approximately 1.729, which, when compared with our t-test statistic of approximately 5.345, leads to the conclusion that the null hypothesis should be rejected.

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Most popular questions from this chapter

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

Suppose that college students with a checking account typically write relatively few checks in any given month, whereas people who are not college students typically write many more checks during a month. Suppose that \(50 \%\) of a bank's accounts are held by students and that \(50 \%\) are held by people who are not college students. Let \(x\) represent the number of checks written in a given month by a randomly selected bank customer. a. Give a sketch of what the population distribution of \(x\) might look like. b. Suppose that the mean value of \(x\) is 22.0 and that the standard deviation is \(16.5 .\) If a random sample of \(n=\) 100 customers is to be selected and \(\bar{x}\) denotes the sample mean number of checks written during a particular month, where is the sampling distribution of \(\bar{x}\) centered, and what is the standard deviation of the sampling distribution of \(\bar{x}\) ? Sketch a rough picture of the sampling distribution. c. What is the approximate probability that \(\bar{x}\) is at most \(20 ?\) At least \(25 ?\)

The authors of the paper "Mean Platelet Volume Could Be Possible Biomarker in Early Diagnosis and Monitoring of Gastric Cancer" (Platelets [2014]: 592-594) wondered if mean platelet volume (MPV) might be a way to distinguish patients with gastric cancer from patients who did not have gastric cancer. MPV was recorded for 31 patients with gastric cancer. The sample mean was reported to be 8.31 femtoliters (fL) and the sample standard deviation was reported to be \(0.78 \mathrm{fL}\). For healthy people, the mean MPV is 7.85 fL. Is there convincing evidence that the mean MPV for patients with gastric cancer is greater than \(7.85 \mathrm{fL} ?\) For purposes of this exercise, you can assume that the sample of 31 patients with gastric cancer is representative of the population of all patients with gastric cancer.

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, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination ("Study Goals and Procrastination Tendencies at Different Stages of the Undergraduate Degree," Studies in Higher Education [2016]: 2028-2043). Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to \(64,\) with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 37.02 and the standard deviation was 6.44 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean procrastination scale for first-year students at this college. (Hint: See Example 12.7.) b. Based on your interval, is 40 a plausible value for the population mean score? What does this imply about the population of first-year students?

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