A multimedia program designed to improve dietary behavior among low-income women was evaluated by comparing women who were randomly assigned to intervention and control groups. The intervention was a...

Extracted text: A multimedia program designed to improve dietary behavior among low-income women was evaluated by comparing women who were randomly assigned to intervention and control groups. The intervention was a 30-minute session in a computer kiosk in the Food Stamp office. One of the outcomes was the score on a knowledge test taken about 2 months after the program. Here is a summary of the data: Mean St.Dev Intervention 166 5.09 1.15 Control 219 4.33 1.16 a) The test had six multiple-choice items that were scored as correct or in- correct, so the total score was an integer between 0 and 6. Do you think this data is Normally distributed? Briefly explain. b) Use an appropriate significance test to determine if the variation differs among the two populations. Use a = .03. c) Carry out the significance test using a two-sided alternative. Report the test statistic with the degrees of freedom and the P-value. d) Find a 87.34% confidence interval for the difference between the two means. Compare the information given by the interval with the information given by the significance test.