Download STA 2023 Summary – Student`s t distribution Sanchez Name: 1. The

STA 2023
Summary – Student’s t distribution
Sanchez
Name: __________________________________________
1. The student’s t distribution has one parameter, degrees of freedom. It is a bell shaped
continuous distribution but below the standard normal distribution. As degrees of
freedom increases the t-distribution approaches the standard normal distribution.
deg rees of freedom    df
Problem 1.
a) Find the area below the standard normal between z=-1 and z=1
Answer: normalcdf(-1, 1)=0.6826894809
b) Find the area below the t-distribution with 10 degrees of freedom between z=-1 and z=1
Answer: tcdf(-1, 1, 10)=0.6591068677
c) Find the area below the t-distribution with 28 degrees of freedom between z=-1 and z=1
Answer: tcdf(-1, 1, 28)=0.6741252928
d) Find the area below the t-distribution with 200 degrees of freedom between z=-1 and z=1
Answer: tcdf(-1, 1, 200)=0.6814811516
e) Find the ordinate (the y-value) of the standard normal probability density function at the center.
Answer: normalpdf(0)=0.3989422804
f) Find the ordinate (the y-value) of the student t probability density function with 4 degrees of
freedom, at the center.
Answer: tpdf(0, 4)=0.375
g) Find the ordinate (the y-value) of the student t probability density function with 10 degrees of
freedom, at the center.
Answer: tpdf(0, 10)=0.389108384
h) Find the ordinate (the y-value) of the student t probability density function with 200 degrees of
freedom, at the center.
Answer: tpdf(0, 200)=0.3984439162
Problem 2. Find the following probabilities
a) P(-1.25 < t < 2.05, df =14 ) = tcdf(-1.25, 2.05, 14) = 0.8543145566
b) P(t < 2.10, df =24) = tcdf(  10 9 , 2.10, 24) = 0.9767892494
c) P(t > 2.10, df =17) = tcdf( 2.10, 10 9 , 17 ) = 0.0254799226
d) P(t < -2.48, df = 34) = tcdf(  10 9 ,  2.48, 34) = 0.0091255827
Inverse problems. For the inverse problem use the EQUATION SOLVER.
Problem. Use the EQUATION SOLVER to find
a) x, if P(t < x. df = 15) = 0.2742
Solution : Eq: 0 =tcdf(  10 9 , x, 15)  0.2742
Answer: x = -0.6140233993
b) x, if P(1.34 < t < x. df = 20 ) = 0.0708
Solution : Eq: 0 =tcdf( 1.34, x, 20)  0.0708
Answer: x = 2.0503098839
ALPHA SOLVE
ALPHA SOLVE
Note: t  is the t-value below which the probability under the t-distribution is
2

2
c) If df = 14, find t 0.05
Solution: Eq: 0 = tcdf ( 10 9 , x, 14)  0.05
Answer: x = -2.046169055
ALPHA SOLVE
d) If df = 25, find t 0.90
Solution: Eq: 0 = tcdf ( 10 9 , x, 25)  0.90
Answer: x = 1.3163450725
ALPHA SOLVE
Using the Student’s t distribution to make inferences about the mean of a populations.
Requirements: to use this method
1. The population under investigation must be normal
2. The standard deviation  of the population is unknown, and the sample standard deviation s is
used in its place.
Problem: A sample of size n = 20 is taken from a normal population. The statistical result of the
sample were a mean x  35.6 and standard deviation s=2.14. Find a 99% confidence interval
for the true mean  of the population.
Solution:
Formula : x  t 
2
s
n
,   1  confidence  1  0.99  0.01
deg rees of freedom  n  1  20  1  19, t   t 0.01  t 0.005  x
2
2
Solve the equation eq : 0  tcdf ( 10 , x, 19)  0.005
Answer : t   t 0.005  x  2.860934604
9
2
x  t
2
s
n
 35.6  2.860934604 
2.14
20
 35.6  1.36901027
34.23098973    36.96901027
Answer : P(34.23098973    36.96901027)  0.99