In: Statistics and Probability
4. The following are 20 observations of response times (in seconds) from a random sample of participants on a cognitive psychology task:
1.7 0.8 4.3 2.9 2.3 1.1 2.2 1.8 2.0 1.2 4.4 1.6 3.8 1.5 2.8
3.3 1.8 2.5 2.7 1.6
(a) Calculate the mean and unbiased standard deviation of the response times.
(b) Construct and interpret a 95% confidence interval for μ, which is the true mean response time. Assume that σ = 1.5.
(c) Construct and interpret a 90% confidence interval for μ, which is the true mean response time. Assume that σ is unknown.
(d) What would happen to the width of your confidence interval if our sample size increased? Explain.
a)
Sample Mean, x̅ = ΣX/n = 2.3150
sample std dev , s = √(Σ(X- x̅ )²/(n-1) )
= 1.0246
b)
Level of Significance , α =
0.05
' ' '
z value= z α/2= 1.960 [Excel
formula =NORMSINV(α/2) ]
Standard Error , SE = σ/√n = 1.5000 /
√ 20 = 0.3354
margin of error, E=Z*SE = 1.9600
* 0.3354 = 0.6574
confidence interval is
Interval Lower Limit = x̅ - E = 2.32
- 0.657392 = 1.6576
Interval Upper Limit = x̅ + E = 2.32
- 0.657392 = 2.9724
95% confidence interval is (
1.6576 < µ < 2.9724
)
c)
Level of Significance , α =
0.1
degree of freedom= DF=n-1= 19
't value=' tα/2= 1.729 [Excel
formula =t.inv(α/2,df) ]
Standard Error , SE = s/√n = 1.0246 /
√ 20 = 0.2291
margin of error , E=t*SE = 1.7291
* 0.2291 = 0.3961
confidence interval is
Interval Lower Limit = x̅ - E = 2.32
- 0.396149 = 1.9189
Interval Upper Limit = x̅ + E = 2.32
- 0.396149 = 2.7111
90% confidence interval is (
1.9189 < µ < 2.7111
)
d)
width of your confidence interval will get decreased if our sample size increased