An Israeli researcher randomly assigned 48 college students to tables of 6 people to go out to eat. Half

 An Israeli researcher randomly assigned 48 college students to tables of 6 people to go out to eat. Half were to pay for their individual meals and half were to split the final check depending on their assigned table and they were told this before ordering. There were a total of 4 tables in each treatment (split or individual) for a grand total of 48 participants at 8 tables.  The cost of each person’s meal was recorded. Included here is the data file, please note we are only using the payment and cost columns.  Please also note the cost is presented in Israeli Shekel.

1. Create a hypothesis test in the correct statistical language. What do you suspect happened and why?
2. Go to StatKey  http://www.lock5stat.com/StatKey/bootstrap_1_quant_1_cat/bootstrap_1_quant_1_cat.html and select “Restaurant Bill” or use the attached file. Looking at the original sample which average bill was higher and by how much?
3. Construct a 95% confidence interval using the randomization (bootstrap) and report a p value for your null hypothesis. Explain how you did it.
4. Repeat this exercise using the formulas. You will need to know that the sample standard deviations are 14.33 and 12.54 for Split and Individual respectively.
5. Draw a conclusion in correct terms of the original problem and explain why this is what you would expect or not.