Question

People tend to be more generous after receiving good news. Are they less generous after receiving bad news? The average tip left by adult Americans is 20%. Give 20 patrons of a restaurant a message on their bill warning them that tomorrow’s weather will be bad and record the tip percentage they leave. Table 1 contains the tips as a percentage of the total bill.

Table 1: Problem 5.3 18.0 19.1 19.2 18.8 18.4 19.0 18.5 16.1 16.8 18.2 14.0 17.0 13.6 17.5 20.0 20.2 18.8 18.0 23.2 19.4

Suppose that tip percentage is Normal with = 2 and assume that the patrons in this study are a random sample of all patrons of this restaurant. Is there good evidence that the mean tip percentage for all patrons of this restaurant is less than 20 when they receive a message warning them of tomorrow’s bad weather? State H0 and Ha and carry out a significance test. Use significance level 0.05.

Answer 1

To test if there is good evidence that the mean tip percentage for all patrons of this restaurant is less than 20 when they receive a message warning them of bad weather, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha) and conduct a significance test.

Step-by-step explanation:

To test if there is good evidence that the mean tip percentage for all patrons of this restaurant is less than 20 when they receive a message warning them of bad weather, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha) and conduct a significance test.

H0: The mean tip percentage for all patrons of this restaurant is >= 20

Ha: The mean tip percentage for all patrons of this restaurant is < 20

Using the given data and assuming a normal distribution with a mean (μ) of 20% and a standard deviation (σ) of 2%, we can perform a one-sample z-test.

With a significance level of 0.05, we can calculate the test statistic:

z = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Plugging in the values from the data provided, we get:

z = (18.0 - 20) / (2 / sqrt(20)) = -1.4142

Using a z-table or calculator, we find that the critical value for a significance level of 0.05 is -1.645 (approximately).

Since the calculated test statistic (-1.4142) is greater than the critical value (-1.645), we do not have enough evidence to reject the null hypothesis. Therefore, there is no significant evidence to suggest that the mean tip percentage for all patrons of this restaurant is less than 20% when they receive the bad weather warning.