null hypothesis Problem #8(b): test statistic (correct to 3 decimals ) Problem #
ID: 3363831 • Letter: N
Question
null hypothesis
Problem #8(b):
test statistic (correct to 3 decimals)
Problem #8(c):
critical value (correct to 3 decimals)
(A) Do not reject H0 since the p-value is equal to 0.0142 which is less than .05. (B) Reject H0 since the p-value is equal to 0.0071 which is less than .05. (C) Do not reject H0 since the absolue value of the answer in (b) is less than the answer in (c). (D) Reject H0 since the absolue value of the answer in (b) is less than the answer in (c). (E) Do not reject H0 since the p-value is equal to 0.0071 which is less than .05. (F) Do not reject H0 since the absolue value of the answer in (b) is greater than the answer in (c). (G) Reject H0 since the p-value is equal to 0.0142 which is less than .05. (H) Reject H0 since the absolue value of the answer in (b) is greater than the answer in (c).
An observational study of Alzheimer's disease (AD) obtained data from 15 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 15 0 75.95 1.68 6.50 77.00 79.25 87.00 92.25 93.00 Control 9 0 64.65 4.29 12.88 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 not 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?
(A) H0:1 2(B) H0:1 2(C) H0:1 2(D) H0:1 = 2(E) H0:1 < 2(F) H0:1 > 2
Problem #8(a): SelectABCDEF
null hypothesis
Problem #8(b):
test statistic (correct to 3 decimals)
Problem #8(c):
critical value (correct to 3 decimals)
(A) Do not reject H0 since the p-value is equal to 0.0142 which is less than .05. (B) Reject H0 since the p-value is equal to 0.0071 which is less than .05. (C) Do not reject H0 since the absolue value of the answer in (b) is less than the answer in (c). (D) Reject H0 since the absolue value of the answer in (b) is less than the answer in (c). (E) Do not reject H0 since the p-value is equal to 0.0071 which is less than .05. (F) Do not reject H0 since the absolue value of the answer in (b) is greater than the answer in (c). (G) Reject H0 since the p-value is equal to 0.0142 which is less than .05. (H) Reject H0 since the absolue value of the answer in (b) is greater than the answer in (c).
Explanation / Answer
Given that,
mean(x)=75.95
standard deviation , 1 =6.5
number(n1)=15
y(mean)=64.95
standard deviation, 2 =12.88
number(n2)=9
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=75.95-64.95/sqrt((42.25/15)+(165.8944/9))
zo =2.39
| zo | =2.39
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =2.386 & | 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 != 2.39 ) = 0.01702
hence value of p0.05 > 0.01702,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 2.39
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.01702
Reject H0 since the p-value is equal to 0.0071 which is less than .05
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