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 = 24 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
= 0.05.Find the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.)
To test whether the mean daily revenue has decreased after the price increase, we can set up the following hypotheses:
Null hypothesis (H0): The mean daily revenue is equal to or greater than its value before the price increase (μ >= 75).
Alternative hypothesis (H1): The mean daily revenue is less than its value before the price increase (μ < 75).
We will use a one-tailed t-test to test these hypotheses.
Given that the sample mean daily revenue is $70.00, the population mean (before the price increase) was $75.00, the sample standard deviation is $4.20, and the sample size is 24, we can calculate the test statistic as follows:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (70 - 75) / (4.20 / sqrt(24))
t ≈ (-5) / (4.20 / 4.898)
t ≈ -5 / 0.858
t ≈ -5.828
Using a t-table or a t-distribution calculator with 23 degrees of freedom (24 - 1), we can find the critical value for a one-tailed test with a significance level of 0.05. In this case, the critical value is approximately -1.713.
Since the t-test statistic (-5.828) is less than the critical value (-1.713), we can reject the null hypothesis. This means that the evidence suggests that the mean daily revenue has decreased from its value before the price increase.
To calculate the p-value, we can use the t-distribution calculator with 23 degrees of freedom and the test statistic (-5.828). The p-value is the probability of observing a test statistic more extreme than the calculated value under the null hypothesis.
The p-value is calculated to be approximately 0.000 (3 decimal places). This indicates strong evidence against the null hypothesis and supports the conclusion that the mean daily revenue has decreased.
In conclusion, based on the given information and test results, there is evidence to suggest that the mean daily revenue has decreased after the price increase.