Problem #8: An observational study of Alzheimer\'s disease (AD) obtained data fr

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Problem #8: An observational study of Alzheimer's disease (AD) obtained data from 13 AD patients exhibiting moderate dementia and selected a group of 6 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*Mean SE Mean StDev Minimum 1 Median 3 Maximum Alzheimers 13 7e.31 Control 1.80 6.49 77.00 79.25 87.00 92.25 93.00 5.24 12.4 54.00 56.00 65.00 82.00 89.00 6 0 66.40 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 not equa (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? Select | null hypothesis Problem #8(a): Problem #8(b): Problem #8(c): test statistic (correct to 3 decimals) critical value (correct to 3 decimals) (A) Reject Ho since the absolue value of the answer in (b) is less than the answer in (c) (B) Do not reject Ho since the absolue value of the answer in (b) is greater than the answer in (c) (C) Reject Ho since the p-value is equal to 0.0316 which is less than 05 (D) Do not reject Ho since the abolue value of the answer in (b) is less than the answer in (c). (E) Do not reject Ho since the p-value is eal to 0.0316 which is less than .05 (F) Reject Ho since the p-value is equal to 0.0158 which is less than.05 (G) Do not reject Ho since the p-value is equal to 0.0158 which is less than.05 (H) Reject Ho since the absolue value of the answer in (b) is greater than the answer in (c) Problem #B(d): Se conclusion Just Save | | Submit Problem #8 for Grading |

Explanation / Answer

Given that,
mean(x)=78.31
standard deviation , 1 =6.49
number(n1)=13
y(mean)=66.4
standard deviation, 2 =12.84
number(n2)=6
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=78.31-66.4/sqrt((42.1201/13)+(164.8656/6))
zo =2.149
| zo | =2.149
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =2.149 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.149 ) = 0.03164
hence value of p0.05 > 0.03164,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 2.149
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.03164
OPTION C:reject H0 since the p-value is equal to 0.0316 which is less than 0.05

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