Download STATISTICS AND THE TI-83

STATISTICS AND THE TI-83
Lesson #9 –Inferential Statistics: ESTIMATION
1. Inference on the Mean of a Population (large sample)
EXERCISE 1. A simple random sample of 36 items has a sample mean of 42. The standard
deviation of the population is =9.
a. What is the point estimate of the population mean?
Answer: 42

9
9
b. What is the standard error of the mean? Answer:  x 

  1.5
n
36 6
c. Provide a 90% confidence interval for the population mean.
Formula: x  z 






2

n
, x  42 ,  = 9, n = 36, =1-0.90 = 0.10, Z0.05  1.645
STAT  TESTS 7 (z Interval) Stats
ENTER
 =9
 x  42
 n = 36
 C-Level: 0.90
answer: ( 39.533, 44.467)
 Calculate ENTER
Conclusion: We are 90% sure that the interval ( 39.533, 44.467) contains the mean.
d. At a 90% probability, what can be said about the margin of error?
Answer: Max E=44.467-42=2.467
e. If we want to lower the margin of error to 2, what should be the size of
2
z    
2
 2

1.645  9 
  55
the sample used? . Use the formula n  
 answer: n  


2
 E 


Exercise 2. A sample of 64 cigarettes of a certain brand was tested for nicotine
content and gave an average of 20 milligrams and a standard deviation of 4 milligrams.
Find a 98% confidence interval for the true mean.
Formula: x  z 
2
s
, x  20 , s = 4, n = 64, =1-0.98 = 0.02, Z0.01  2.326
n
Note: we are using s as an estimate for 


STAT
TESTS 7 Zinterval
 Calculate ENTER
stats ENTER  4  20  64  .98
answer: (18.837, 21.163)
At a 98% probability the margin of error is 21.163-20 = 1.163
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Exercise 3. In exercise 2, if we want to lower the margin of error 1 mg, what should
the sample size be changed to?
2
2.326  4 
 = 87
E=1, use  = 4, z 0.01 =2.326, n = 


1
Ans.: 87
II. The student’s t distribution
1. tpdf (value of t, df) computes the value of the student’s t distribution function at
a given value, for a specified number of degrees of freedom.
The maximum or peak of the curve occurs at mean = 0.
Exercise 4. Find the ordinate of the t-distribution function with 12 degrees of
freedom for t=1.28. Find the ordinate value at the peak of the curve.


2nd DIST 4 tpdc(1.28, 12) ENTER
answer: .1700546438
2nd ENTRY (change 1.28 to 0) ENTER answer: tpdf(0, 12) = .3907263052
2. tcdf (lower, upper, df) computes the Student-t distribution probability between
lowerbound and upperbound values for the specified df (degrees of freedom)
Exercise 5. Compute P(-2 < t < 1, df = 15)

2nd DISTR 5 tcdf (-2, 1, 15) ENTER
answer: .8014424299
Exercise 6. Compute P(-∞ < t < -1.943, df = 6)

2nd DISTR 5 tcdf (-10^9, -1.943, 6) ENTER answer: .0500124985
Exercise 7. Use DRAW to compute P(t >- 2.552, df = 18)

WINDOW

2nd DISTR
Xmin=-5
Xmax=5
Ymin=-.05
Ymax=2nd DIST 4 tpdf (0, 18) ENTER
DRAW 2 Shade ENTER t(-2.552, 10^9, 18)
answer: .989992
Using MATH SOLVER to find INVt-value.
Exercise 7. For 10 degrees of freedom, find the value of t below which we find an area of
0.05, that is, find t 0.05, 10


MATH 0  CLEAR (to clear any existing equation)
Eqn: 0=2nd DIST 5 tcdf(-10^9, x, 10) – 0.05 ENTER ALPHA SOLVE
answer: t=-2.978723404
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III. Inference on the Mean of a Population (small sample)
Exercise 8. A group of 20 people lost an average of 5 pounds a week with a standard
deviation of 1.3 pounds, by going through some special dieting process. Assuming that
the weight lost is a normal distribution, find a 95% confidence interval for the true
average weight loss for people who go through this dieting process. What is the
maximum error of estimation?
Note: this is a t-interval because  is unknown, n is small and the population is
assumed to be normal.
Formula: x  t 
2
df  19, t 0.025

STAT
s
, df  n  1 , x  5 , s = 1.3, n = 20, =1-0.95 = 0.05,
n
 2.093
TESTS 8 T interval
stats ENTER  5  1.3  20  0.95
  Calculate ENTER
answer: (4.3916, 5.6084)
At a 95% probability the margin of error is = 5.6084 - 5= 0.6084
Exercise 9. A set of 12 experimental animals was fed a special diet for 3 weeks and
produced the following gains in weight: 30, 22, 32, 26, 24, 40, 34, 36, 32, 33, 28, 30
pounds. Find a 90% confidence interval for the population mean gain in weight,
assuming that gain in weight is a normal variable.
 STAT 1 EDIT
(use  to move to an available column)
 2nd INS Name = W
ENTER
  30 ENTER 22 ENTER 32 ENTER 26 ENTER 24 ENTER 40 ENTER
34 ENTER 36 ENTER 32 ENTER 33 ENTER 28 ENTER 30 ENTER

STAT


ENTER Freq: 1   C-Level .90
 Calculate ENTER
 TESTS
8 Tinterval DATA
ENTER  LIST 2ndLIST W
answer: (27.945, 33.222)
x = 30.5833333 Sx=5.089353117
Note: you could use {30, 22, 32, 26, 24, 40, 34, 36, 32, 33, 28, 30} STO  W to create
the list W.
Exercise 10. Find a 90% confidence interval for the mean of a normal population from which
the sample {3, 7, 1, 8, 5, 3, 4} was taken.



{3, 7, 1, 8, 5, 3, 4} STO  L1 ENTER
STAT TESTS 8 Tinterval DATA
  C-LEVEL .90  Calculate ENTER
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ENTER 
LIST 2nd L1
Answer: (2.6367. 6.2205)
IV. Inference: proportion of a population
Exercise 11. A sample of 80 adults was interviewed and 50 of them indicated that they favor
the Republican candidate for state governor.
a. What is the point estimate for the true proportion of adults who favor the Republican candidate
for state governor?
answer:
p
50
 .625 or 62.5%
80
 5080 ENTER
answer: 0.625
b. Estimate the standard error of the distribution of sample proportions pˆ
pq
0.625  0.375

 .0541265877
n
80

(0.625  0.375  80) ENTER
answer: .0541265877
c. Provide a 90% confidence interval for the population proportion p
answer:  pˆ 
ˆ  z
Formula: p
2
pˆ qˆ
, pˆ =0.625, qˆ = 1 - pˆ =1 -0.625 = 0.375, n=80,
n
=1 -0.90= 0.10, z0.05  1.645
 STAT  TESTS  A (1-Prop z Int) ENTER
 x = 50
  n=80
  C-Level: 0.90
  Calculate ENTER
Ans.: (0.53597, 0.71403) pˆ =0.625, n=80
Conclusion: we are 90% sure that the interval : (0.53597, 0.71403) contains p.
d) At the 90% probability, what can be said about the margin of error?
answer : Max E = 0.71043-0.625 = 0.08543
e) If we want to lower the marginal error to 5%, what should be the sample size used?
2
answer:

 
z   pq
2
 
1.645 0.6250.375
 2 
n

 254
E
0.05 2


1.645^2  0.625  0.375  (0.05^2)
answer: 254
Exercise 12. Find a 98% confidence interval for the proportion of college students who
favor capital punishment, if a sample of 100 students indicates that 80 of them are in
favor.
 STAT
TESTS 2nd ALPHA A PropZInt


 x: 80  n:100  C-level: .98
ENTER
answer: (0.70695, 0.89305)
 Calculate
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V. PROGRAM . Create a program for computing the sample size needed in interval estimation:
mean and proportions.
PRGM  NEW Create New ENTER
Name = NSIZE
PROGRAM: NSIZE
: Prompt C, E, S, P
ENTER (use p=0 for a mean problem and s=0
for a proportion problem
: (1+C)/2 STO  A ENTER
: 2nd DIST invNorm(A)
STO  Z ENTER
: PRGM CTL 9 Lbl L
ENTER
: PRGM 1 If S = 0
ENTER
: PRGM 0 GoTo M
ENTER
: (Z*S/E)^2
STO  N ENTER
: GoTo F
ENTER
: Lbl M
ENTER
: z^2*P*(1-P)/E
STO  N ENTER
: Lbl F
ENTER
: MATH Num 5 int(N)
STO N ENTER
: Disp “SAMPLE SIZE NEEDED IS”, N ENTER
Exercise 13. We wish to determine the average length of time that an automobile is
parked on the campus parking lot. How large a sample is needed, in order to make a
1
statement with a 90% confidence that our mean is within
hour of the true mean.
2
Assume the population standard deviation is 2 hours.
Use the program NSIZE





PRGM NSIZE ENTER
C=? .90 ENTER
E=? .5 ENTER
S=? 2 ENTER
P=? 0 ENTER
answer: SAMPLE SIZE NEEDED IS 43
Exercise 14. A manufacturer of parts believes that approximately 5% of his products
are defective. If he wishes to estimate the true proportion to within 0.005 and to be
certain with a probability of 99% of being correct, how large a sample should he take?
Use the program NSIZE
 PRGM NSIZE ENTER
 C=? .99 ENTER
 E=? .005 ENTER
 S=? 0
ENTER
 P=? 0.05 ENTER
answer: sample size needed is 12606 DONE
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