Question

A student organization uses the proceeds from a particular soft-drink dispensing machine to finance its activities. The price per can had been \(\$ 0.75\) for a long time, and the average daily revenue during that period had been \(\$ 75.00\). The price was recently increased to \(\$ 1.00\) per can. A random sample of \(n=20\) days after the price increase yielded a sample average daily revenue and sample standard deviation of \(\$ 70.00\) and \(\$ 4.20\), respectively. Does this information suggest that the true average daily revenue has decreased from its value before the price increase? Test the appropriate hypotheses using \(\alpha=.05\).

Short answer

The short answer would depend on the calculated p-value and t statistic. However, since these calculations have not been carried out in this instance, we can not provide a definitive answer. Depending on the results, it could be that the data does not provide strong evidence against the null hypothesis, in which case we wouldn't reject it, or it might be the case that there is strong evidence against the null hypothesis, leading to its rejection in favor of the alternative.

Step-by-step solution

Formulate the hypotheses

The null hypothesis \(H_0\) states that the mean revenue has not decreased, it is still $75. So, \(H_0: \mu = 75\). The alternative hypothesis \(H_1\) states that the mean revenue has decreased. So, \(H_1: \mu < 75\).

Establish the significance level

The significance level is given as \(\alpha = 0.05\). This is the probability of rejecting the null hypothesis when it is true.

Calculate the Test Statistic

For the sample, the mean (\(\bar{x}\)) is $70, the sample standard deviation (s) is $4.20, and the sample size (n) is 20. We use the formula for the test statistic in a t test, which is \(t = (\bar{x} - \mu) / (s/ \sqrt{n})\). Substituting the given values we get \(t = (70 - 75) / (4.20 / \sqrt{20})\).

Calculate the p-value

After calculating the test statistic we look up this value in the t-distribution table (with n-1 degrees of freedom, which is 19) to find the p-value. This p-value tells us the probability of obtaining a result as extreme as our test statistic assuming the null hypothesis is true.

Conclude the hypothesis test

We then compare the p-value to our significance level (\(\alpha\)) to decide whether to reject the null hypothesis. If the p-value is less than or equal to \(\alpha\), we reject the null hypothesis in favor of the alternative. Otherwise, we do not have enough evidence to reject the null hypothesis.