Question

A sample of Alzheimer's patients is tested to assess the amount of time in stage IV sleep these patients get in a 24-hour period. Number of minutes spent in Stage IV sleep is recorded for 61 patients. The mean stage IV sleep over a 24 hour period of time for these 61 patients was 48 minutes with a standard deviation of 14 minutes. (a) Compute 90% confidence interval for mean stage IV sleep. Interpret this confidence interval. (b) It has been believed that individuals suffering from Alzheimer's Disease may spend less time per night in the deeper stages of sleep. Test the hypothesis at 10% significance level if the true mean stage IV sleep of Alzheimer patients is less than 49 minutes. Provide the hypothesis statement Calculate the test statistic value Determine the probability value (c) Could the confidence interval in part (a) be used to test the hypothesis in part (b)? Why or why not? (1 mark)

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =48
standard deviation, s =14
sample size, n =61
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 14/ sqrt ( 61) )
= 1.793
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 60 d.f is 1.671
margin of error = 1.671 * 1.793
= 2.995
III.
CI = x ± margin of error
confidence interval = [ 48 ± 2.995 ]
= [ 45.005 , 50.995 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =48
standard deviation, s =14
sample size, n =61
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 60 d.f is 1.671
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 48 ± t a/2 ( 14/ Sqrt ( 61) ]
= [ 48-(1.671 * 1.793) , 48+(1.671 * 1.793) ]
= [ 45.005 , 50.995 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 45.005 , 50.995 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
b.
Given that,
population mean(u)=49
sample mean, x =48
standard deviation, s =14
number (n)=61
null, Ho: μ=49
alternate, H1: μ<49
level of significance, α = 0.1
from standard normal table,left tailed t α/2 =1.296
since our test is left-tailed
reject Ho, if to < -1.296
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =48-49/(14/sqrt(61))
to =-0.5579
| to | =0.5579
critical value
the value of |t α| with n-1 = 60 d.f is 1.296
we got |to| =0.5579 & | t α | =1.296
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.5579 ) = 0.2895
hence value of p0.1 < 0.2895,here we do not reject Ho
ANSWERS
---------------

T test for single mean with known sample standard deviation
null, Ho: μ=49
alternate, H1: μ<49
test statistic: -0.5579
critical value: -1.296
decision: do not reject Ho
p-value: 0.2895
we do not have enough evidence to support the claim that if the true mean stage IV sleep of Alzheimer patients is less than 49 minutes.