Question

Olestra is a fat substitute approved by the FDA for use in snack foods. The article Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips (Journal of the American Medical Association, 1998: 150-152) reported that in an experiment, 17.6% of the 529 individuals in the TG control group experience an adverse GI event, whereas 15.8% of the 563 individuals in the olestra treatment group experienced such an event. Does it appear that the incident rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group? Using a α{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3B1;</mi></math>"}= 0.01, carry out a test of significance by answering the questions below in as much detail as possible. [Hint: This is a two-sample proportion problem.] Use this information to answer questions 9 to 13.

A) What are the null and alternative hypotheses for the test?

	A) P2 0 vs
	B) vs
	C) vs
	D) vs
	E) vs

B) Compute the test statistic and degrees of freedom (if applicable).

	A) 0.5
	B) 0.7967
	C) 0.0226
	D) 0.018
	E) 1.0871

C) What is the p-value of the test?

	A) 0.7881
	B) 0.2119
	C) 0.4438
	D) 0.1379
	E) 0.2757

D) What is your decision and why?

	A) Reject since p-value < 0.01
	B) Fail to reject since p-value < 0.01
	C) Reject since p-value > 0.05
	D) Reject since p-value > 0.01
	E) Fail to reject since p-value >0.01

E) What is your conclusion in the context of the problem?

	A) We have evidence to show that there is a difference in incidence rate between the two groups.
	B) We do not have evidence to show that there is a difference in incidence rate between the two groups.
	C) We have evidence to show that the incidence rate of control group is greater than the incidence rate of treatment group.
	D) We do not have evidence to show that the incidence rate of control group is greater than the incidence rate of treatment group.
	E) We have evidence to show that the incidence rate of control group is less than the incidence rate of treatment group.

Answer 1

From given information, n1 529 x1 = 93 n2 563 x2 89 a) The following null and alternative hypotheses need to be tested: Ho: P1 P2 0 versus Ha: P1- P2 0 b) Test statistics: Р — Ра = V(1-p" + T12 I1+2 where p* Here P1 0.176 P2 0.158 The value of the pooled proportion is computed as p* (93+89)/ (529 563) P*0.1667 Z (0.176- 0.158) V 0.1667*(1 - 0.1667) ((1/529) (1/563)) 0.7976 = Test statistics Z 0.7976 c) P value P(Z > 0.7976) = 2*(1 - NORM.S.DIST(0.7976,TRUE)) = 0.4251 The P value is 0.4251 d) Since P -value 0.4251 > a = 0.01 so we failed to reject null hypothesis. e) We can conclude that we do not have evidence to show that there is a difference in incidence rate between the two groups.