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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
Based on the t-test, it can be concluded that the mean daily revenue has decreased after increasing the price of the soft drinks, rejecting the null hypothesis at the 0.05 significance level.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (H0) is that the mean daily revenue has not decreased. So, \(H0: \mu \geq \$75.00\).\nThe alternative hypothesis (Ha) is that the mean daily revenue has decreased. So, \(Ha: \mu < \$75.00\).
02

Compute the Test Statistic

The t-value is computed using the formula: \(t = \frac{(\bar{x} - \mu_0)}{(s / \sqrt{n})}\) where \(\bar{x}\) is the sample mean (\$70.00), \( \mu_0\) is the hypothesized population mean (\$75.00), s is the sample standard deviation (\$4.20) and n is the sample size (20).\nPlugging these values in gives: \[t = \frac{(\$70.00 - \$75.00)}{(\$4.20 / \sqrt{20})} = -4.76.\]
03

Find the Critical Value

Since it is a one-tailed test to the left, and with \(\alpha = 0.05\) and degrees of freedom = n - 1 = 20 - 1 = 19, from t-table, the critical value of t (t_critical) for a one-tailed test is -1.729.
04

Making the Decision

Since the calculated t-value (-4.76) is less than the critical t-value (-1.729), we reject the null hypothesis in favor of the alternative hypothesis.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there's a significant difference between the means of two groups. It's particularly helpful when working with small sample sizes and unknown population variances. There are different variations of the t-test, such as one-sample, independent, and paired t-tests. In this exercise, a one-sample t-test is used to compare a single sample mean against a known population mean. This helps to assess whether the new sample mean of daily revenue after the price increase significantly differs from the past revenue mean of $75.00.

In a t-test, you'll first set up two hypotheses:
  • The null hypothesis ( $H_0 $), which suggests no effect or difference (e.g., mean revenue has not decreased).
  • The alternative hypothesis ( $H_a $), which suggests a potential difference (e.g., mean revenue has decreased).
Next, you'll calculate the t-statistic using the sample data, which helps you decide whether to reject the null hypothesis. This process involves comparing the calculated t-value to a critical value derived from a t-distribution table.
sample mean
The sample mean, often denoted as \(\bar{x}\), is the average of all values in a sample. It is calculated by adding up all the data points and dividing by the number of data points in the sample. In our context, the sample mean of the daily revenue collected over 20 days after the price change is $70.00.

The sample mean plays a key role in hypothesis testing because it serves as the best estimate of the true population mean when individual population data is not available. When comparing the sample mean to the known population mean, you can determine if there is a statistically significant difference. A sample mean that significantly deviates from the hypothesized population mean provides evidence to reject the null hypothesis.
standard deviation
Standard deviation measures the amount of variability or dispersion in a set of data. It's a crucial concept because it gives context to the mean, showing how spread out the sample values are around the mean. In hypothesis testing, the sample standard deviation is used to understand the variability within the data set.

In our example, the sample standard deviation is $4.20. This value tells us how much the daily revenue fluctuates around the mean of $70.00 over the sample period. Lower standard deviation indicates that data points are closer to the mean, while a larger standard deviation signifies more dispersion. Standard deviation is also used to calculate the standard error, which is crucial for determining the t-statistic, allowing you to measure how "standard" the sample mean is compared to the hypothesized mean.
critical value
The critical value is a threshold in hypothesis testing that helps determine whether to accept or reject the null hypothesis. It is derived from a probability distribution—in this case, the t-distribution—based on a desired level of significance, denoted as \(\alpha\).

For our test, with a chosen significance level \(\alpha = 0.05\), and 19 degrees of freedom (calculated as the sample size minus one), we find a critical t-value of -1.729. This value is the cutoff point for accepting or rejecting the null hypothesis.

When the calculated t-statistic is more extreme than the critical value, it indicates that the observed data is unlikely under the null hypothesis. In our example, since the calculated t-value of -4.76 is less than -1.729, it falls into the rejection region. Thus, we reject the null hypothesis and conclude that the mean daily revenue probably decreased after the price increase.

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Most popular questions from this chapter

In a study of academic procrastination, the authors of the paper "Correlates and Consequences of Behavioral Procrastination" (Procrastination, Current Issues and New Directions [2000]) reported that for a sample of 411 undergraduate students at a mid-size public university, the mean time spent studying for the final exam in an introductory psychology course was 7.74 hours and the standard deviation of study times was 3.40 hours. Assume that this sample is representative of students taking introductory psychology at this university. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean time spent studying for the introductory psychology final exam. (Hint: See Example 12.7)

Explain the difference between \(\bar{x}\) and \(\mu_{\bar{x}}\)

Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the alcohol content of each bottle is determined. Let \(\mu\) denote the mean alcohol content for the population of all bottles of this brand. Suppose that this sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of (7.8,9.4) a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 7.8 and 9.4 . Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a random sample of size 50 and then calculating the corresponding \(95 \%\) confidence interval is repeated 100 times, exactly 95 of the resulting intervals will include \(\mu .\) Is this statement correct? Why or why not?

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\)

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica [1991]: \(63-67)\). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9\) minutes. Suppose that the standard deviation was \(s=7.7\) minutes. Is there convincing evidence that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes? What assumptions must be reasonable for the one-sample \(t\) test to be appropriate?

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