Question

The data (below) represent a random sample of 9 scores (out of 100) on a midterm exam.
70 90 70 50 40 80 30 100 90
a. Find the sample mean and sample standard deviation.
b. Find the standard error of x bar
c. Assuming that the population distribution of scores is normal, find a 95% confidence interval
for the true mean score.
d. Interpret your result in (c).
e. One claims that the true mean score is 90. Do you reject this claim? Explain.

Answer 1

Part a)

Mean X̅ = Σ Xi / n
X̅ = 620 / 9 = 68.8889

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 4688.8885 / 9 -1 ) = 24.2097

Part b)

Standard Error = S/√(n) = 8.0699

part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 9- 1 ) = 2.306
68.8889 ± t(0.05/2, 9 -1) * 24.2097/√(9)
Lower Limit = 68.8889 - t(0.05/2, 9 -1) 24.2097/√(9)
Lower Limit = 50.2797
Upper Limit = 68.8889 + t(0.05/2, 9 -1) 24.2097/√(9)
Upper Limit = 87.4981
95% Confidence interval is ( 50.2797 , 87.4981 )

Part d)

To Test :-

H0 :- µ = 90
H1 :- µ ≠ 90

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 68.8889 - 90 ) / ( 24.2097 / √(9) )
t = -2.616

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.05 /2, 9-1) = 2.306
| t | > t(α/2, n-1) = 2.616 > 2.306
Result :- Reject null hypothesis

Yes, we reject the claim.