To apply the two-sample t procedures, use Option 1 if you have technology that i

Question

To apply the two-sample t procedures, use Option 1 if you have technology that implements that method. Otherwise, use Option 2. Researchers gave 40 index cards to a waitress at an Italian restaurant in New Jersey. Before delivering the bill to each customer, the waitress randomly selected a card and wrote on the bill the same message that was printed on the index card. Twenty of the cards had the message "The weather is supposed to be really good tomorrow. I hope you enjoy the day! Another 20 cards contained the message "The weather is supposed to be not so good tomorrow. I hope you enjoy the day anyway!" After the customers left, the waitress recorded the amount of the tip (percent of bill) before taxes. Here are the tips for those receiving the good-weather message. 21.3 18.9 19.9 21.1 22.3 23.9 23.2 25.0 22.7 20.5 25.1 22.5 27.4 20.5 22.3 24.4 21.4 22.5 22.2 23.2 The tips for the 20 customers who received the bad-weather message are below 18.3 19.2 19.3 18.8 18.8 19.0 18.8 16.517.2 14.3 17.1 13.7 17.5 20.1 20.4 19.2 18.1 23.5 18.3 19.5 Give a 95% confidence interval for the difference between the mean percent tips for the two different messages. (Round your answers to two decimal places.) to percent You may need to use the appropriate Appendix Table to answer this question.

Explanation / Answer

For the good weather message:

Sample mean, m1 = 22.515

Sample size, n1 = 20

Sample SD, S1 = 2.001

For the bad weather message:

Sample mean, m2 = 18.38

Sample size, n2 = 20

Sample SD, S2 = 2.097

Standard error, SE = (S12/n1 + S22/n2)0.5 = (2.0012/20 + 2.0972/20)0.5 = 0.648

Margin of error, ME = SE*tcritical

For a two tailed t-test with 95% CI and 19 degrees of freedom, tcritical = 2.093

So,

ME = 0.648*2.093 = 1.356

Sample mean difference = m1-m2 = 22.515-18.38 = 4.135

So, required CI is:

4.135-ME to 4.135+ME, which equals

4.135-1.356 to 4.135+1.356, which equals

2.779% to 5.491%

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