Question

A random sample of individuals is selected from a population with μ , and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be with . How much difference is there between the mean for the treated sample and the mean for the original population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.) If there is no treatment effect, how much difference is expected between the sample mean and its population mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denominator of the t statistic.) Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with .

Answer 1

Answer:

A) THE MEAN DIFFERENCE = 22.2 - 20 = 2.2

B) HERE WE NEED TO CALCULATE THE MARGIN OF ERROR

AS GIVEN THE VARIANCE = 384

THEREFORE STANDARD DEVIATION = (384)^(1/2) = 19.59

THE MARGIN OF ERROR = STANDARD DEVIATION / SQRT(N) = 19.59/SQRT(25) = 19.59/5 = 3.91

C) AS THE SAMPLE SIZE IS LESS THEN 30 THEREFORE WE WILL DO IT BY THE T TEST

We need to test the null H0 : µ = 20 against the TWO-sided alternative H1 : µ\neq 20, at level α = 0.05. Since n is SMALL, we will do a SMALL-sample T-test. The rejection region is T >Tα = 1.96OR T<-Tα = -1.96, using the normal table.

T= (X − µ0) /(S/√ n) = (22.2 − 20)/( 19.59/ √ 25) = 0.56 Since T = 0.56 < 1.96, H0 is ACCEPTED. Thus, there is significant evidence at 5% signifi- cance level .

Step-by-step explanation: