Question

Suppose that a pharmaceutical company is interested in testing their new cholesterol medication and recruited 11 sets of identical twins with high blood pressures. For each twin, one was given the new drug and the other was given the old drug, then the reduction in LDL cholesterol level (unit: mg/dL) was measured after 6 months. Summary statistics of the amount of reduction are shown below.

NOTE: Not all outputs given above are useful

	n	x	s
New Drug	11	42	20
Old Drug	11	27	17
Difference (New Drug - Old Drug)	11	15	26

Answer the following questions. If necessary, round your answer to four decimal places.

(a) If the company wishes to test if the average reduction in LDL cholesterol level is higher for the new drug, what would be the most appropriate test to conduct?

The paired t-test for a difference in meansThe two-sample t-test (unequal variances)    The one-sample t-test for a population mean

(b) Conduct the appropriate test. Use the significance level α = 0.05.

STEP 1: Set up null and alternative hypotheses using statistical terms only.

H₀:  --- μD μ1 - μ2 μ  --- > < = ≠
H_A:  --- μ1 - μ2 μD μ  --- < = > ≠

1. μ_D represents the population mean difference between reduction from the new drug and the new old drug (New Drug - Old Drug).
2. μ₁ represents the population mean reduction from the new drug
3. μ₂ represents the population mean reduction from the old drug.

STEP 2: Check all conditions. Make sure your answer is in context of the problem.

STEP 3: Calculate the test-statistic and p-value. If necessary, round your answers to four decimal places.

df =
t =
< p-value <

STEP 4: State the conclusion in statistical terms.

Fail to reject the null hypothesis. Reject the null hypothesis.    

State the conclusion in problem context.

Answer 1

a)The paired t-test for a difference in means

b)

null hypothesis: HO: μd		=	0
Alternate Hypothesis: Ha: μd		>	0

STEP 2: the samples are dependent and randomly selected, we need to assume that difference is normally distributed since sample size is less than 30

STEP 3:

population mean μ=	0
sample mean 'x̄=	15.000
sample size   n=	11.00
sample std deviation s=	26.000
std error 'sx=s/√n=	7.8393
df =n- 1 =10
test stat t ='(x-μ)*√n/sx=	1.9134

0.025 < p value <0.05

STEP 4: since p value <0.05

Reject the null hypothesis.