Question

An observational study of Alzheimer's disease (AD) obtained data from 10 AD patients exhibiting moderate dementia and selected a group of 9 control individuals without AD. AD is a progressive neurodegenerative disease of the elderly and advancing age is known to be a primary risk factor in AD diagnosis. Therefore, it was crucial for the study's credibility to examine whether the ages in the AD group might be significantly different than in the control group. The ages of the subjects in years are summarized in the Minitab Output below. ----------------------------------------------------------------------------------------------------------------------- Descriptive Statistics: Alzheimers, Control Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum Alzheimers 10 0 75.49 2.15 6.81 77.00 79.25 87.00 92.25 93.00 Control 9 0 64.68 4.28 12.85 54.00 56.00 65.00 82.00 89.00 ----------------------------------------------------------------------------------------------------------------------- We want to test if the average age in the Alzheimer's group is significantly different than the control group. Assume that the population variances are equal. (a) What is the null hypothesis? (b) Find the value of the test statistic. (c) Find the 5% critical value. (d) What is the conclusion of the hypothesis test?

for part a: (A) H0 : μ1 ≠ μ2 (B) H0 : μ1 > μ2 (C) H0 : μ1 ≤ μ2 (D) H0 : μ1 = μ2 (E) H0 : μ1 ≥ μ2 (F) H0 : μ1 < μ2 for part d: (A) Reject H0 since the p-value is equal to 0.0099 which is less than .05. (B) Do not reject H0 since the absolue value of the answer in (b) is less than the answer in (c). (C) Do not reject H0 since the absolue value of the answer in (b) is greater than the answer in (c). (D) Do not reject H0 since the p-value is equal to 0.0198 which is less than .05 E) Reject H0 since the p-value is equal to 0.0198 which is less than .05. (F) Reject H0 since the absolue value of the answer in (b) is greater than the answer in (c). G) Do not reject H0 since the p-value is equal to 0.0099 which is less than .05. (H) Reject H0 since the absolue value of the answer in (b) is less than the answer in (c).

Answer 1

a)

(D) H0 : μ1 = μ2

b)

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   75.490                  
standard deviation of sample 1,   s1 =    6.8100                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   64.680                  
standard deviation of sample 2,   s2 =    12.8500                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    75.4900   -   64.7   =   10.810  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    10.1122                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.6462                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   10.8100   -   0   ) /    4.65   =   2.3266

c)

Degree of freedom, DF=   n1+n2-2 =    17  
t-critical value , t* = ± 2.110   (excel formula =t.inv(α/2,df)

d)

p-value =        0.0326

(F) Reject H₀ since the absolute value of the answer in (b) is greater than the answer in (c).

please revert for doubt