Question

2. (30)A researcher is interested in determining if medication A is able to lower patients cholesterol values. The data used is below:

Pre-treatment	Post-treatment
99	95
120	111
97	97
130	132
148	144
122	100
131	120
109	110
140	131
153	154
131	105
120	119
114	107
110	101
116	118

The researcher computed the following results:

            data: pretreatment and posttreatment

t = -2.9305, df = 14, p-value = 0.01096

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval: -11.084051 -1.715949

1. 1. What test was completed?
2. 2. What was the null and alternative hypothesis?
3. 3. What is the interpretation of the Confidence Interval?
4. 4. State your conclusion.

Answer 1

Ans:

a)Paired t test(repeated measures/dependent samples t test)

b)

c)

We are 95% confident that true mean difference falls between -11.084051 and -1.715949

As,both limits are negative,so confidence interval does not include 0 within its limits,we reject the null hypothesis.

d)There is sufficient evidence to conclude that true difference in means is not equal to 0