In: Statistics and Probability
A clinical trial was conducted to test the effectiveness of the drug Zopiclone for treating insomnia in older subjects. Before treatment with Zopiclone, 16 subjects had a mean wake time of 102.8 minutes. After treatment with Zopiclone, the 16 subjects had a mean wake time of 98.9 minutes and a standard deviation of 42.3 minutes.
a. Assume that the 16 samples values appear to be from a normally distributed population, and test the claim that after treatment with Zopiclone, subjects have a mean wake time of less than 102.8 minutes.
b. Test the claim by constructing an appropriate confidence interval.
c. What are the null and alternative hypotheses?
d. What is the value of α= significance level?
e. Is the test two-tailed, left-tailed or right-tailed?
f. What is the test statistic and what is its distribution?
g. What is the value of the test statistic?
h. What is the P-value?
i. What is (are) the critical value(s)?
j. How would you state a conclusion that addresses the original claim?
RE: COMMENT
The value of alpha is stated, does this question contain a typo?
b)
Level of Significance , α =
0.05
degree of freedom= DF=n-1= 15
't value=' tα/2= 2.131 [Excel
formula =t.inv(α/2,df) ]
Standard Error , SE = s/√n = 42.3/√16=
10.5750
margin of error , E=t*SE = 2.131*10.575=
22.5401
confidence interval is
Interval Lower Limit = x̅ - E =
98.9-22.5401= 76.3599
Interval Upper Limit = x̅ + E =
98.9+22.5401= 121.4401
95% Confidence interval is 76.36 < µ
< 121.44 )
c)
Ho : µ = 102.8
Ha : µ < 102.8 (Left tail
test)
d) α=0.05
e) one tail test
f) t distributon
g) Standard Error , SE = s/√n =
42.3/√16= 10.5750
t-test statistic= (x̅ - µ )/SE =
(98.9-102.8)/10.575= -0.3688
h) p-Value = 0.3587
i) critical t value, t* = -1.7531
j) Decision: p-value>α, Do not reject null hypothesis
Conclusion: There is not enough evidence to conclude
that after treatment with Zopiclone,
subjects have a mean wake time of less than 102.8
minutes.