Question

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

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

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, \(Ha: \mu < \$75.00\).

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: \[t = \frac{(\$70.00 - \$75.00)}{(\$4.20 / \sqrt{20})} = -4.76.\]

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.

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.