A researcher has developed a new test to be used in diagnosis of early-stage dem

Question

A researcher has developed a new test to be used in diagnosis of early-stage dementia. The test involves a battery of cognitive tests, and is scored on a 100 point scale, with 100 meaning perfect cognitive function. The test is administered to 52 controls subjects, and 48 subjects who are known to have dementia. Then, the distributions of the two groups will be compared to see if the really can distinguish them.

Suppose the distribution of scores for the control subjects is N(87, 16) (a normal distribution with mean=87 and variance=16). The distribution for patients N(65, 12). Assume a standard significance level of 0.05. Do you think there is enough evidence to reject the null hypothesis? Why or why not? You do not need to calculate a p-value.

Explanation / Answer

Given that,
mean(x)=87
standard deviation , 1 =4
number(n1)=52
y(mean)=65
standard deviation, 2 =3.464
number(n2)=48
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=87-65/sqrt((16/52)+(11.9993/48))
zo =29.4599
| zo | =29.4599
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =29.46 & | 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 != 29.4599 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 29.4599
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim

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