STATISTICS AND THE TI-84
Practice Problems 10 : answers
Hypothesis Testing
1. The average height of females in a college class of a certain college has been 162 centimeters
with a standard deviation of 7 centimeters. Is there reason to believe that there has been a
change in the average height if a random sample of 49 females in the present freshman class has
an average height of 166 centimeters. Use a 0.01 level of significance.
Step 1. State Ho: _
=162_____ State H1:
_
≠162 ___________________
Step 2. Type of test; _two- tailed test_
Step 3. Level of significance: _
=0.01__
Step 4. critical value of the statistic: _ z=± 2.576____
 2nd DIST
3 invNormal (0.005) answer: -2.575829303
Step 5. Diagram
0.005
0.005
z=-2.576
  162
z=2.576
Step 6. Decision rule: _Reject H0 if z computed from evidence is <-2.576 or >2.576
or if p-value<0.01
Step 7. Compute the statistic.: __z=4________________
Evidence: n=49,
x  166
z
x

n
 7

166  162 34

 4  2.576. Therefore Re ject H 0
7
1
49
Compute the p-value
 2 2nd DIST 2 normalcdf(4, EE5) ENTER
or
 2 2nd DIST 2 normalcdf(166, 1EE5, 162, 1 ) ENTER
p-value= 2P(z>4)=2(0.000031686)=0.0000634 <0.01 (Using the TI-83)
p-value is approximately 0 <0.01
Step 8. Conclusion: Reject H0. We have statistical evidence at a 0.01 level of significance to
believe that there has been a change in the average height of females in the present freshman
class
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2. A random sample of 9 cigarettes of a certain brand has an average nicotine content of 4.2
milligrams and a standard deviation of 1.5 milligrams. Is this in line with the manufacturer's claim
that the average nicotine content does not exceed 3.4 milligrams? Use a 0.01 level of
significance and assume the distribution of nicotine contents to be normal. Show all the required
steps and give a precise conclusion.
Step 1. State Ho: _ ≤3.4________________________
State H1: _ >3.4 ____________________________
Step 2. Type of test; _right- tailed test_________
Step 3. Level of significance: _
=0.01__________
Step 4. critical value of the statistic: _
t=2.896 (8 degrees of freedom) ____
 MATH 0  CLEAR
 eqn: 0= 2nd DISTR 5
 eqn: 0=tcdf-EE5, x, 8) - 0.99 ENTER ALPHA SOLVE
Step 5. Diagram
answer: 2.896459446
0.01
t=2.896
  3.4
Step 6. Decision rule: _Reject H0 if t computed from evidence is >2.896
or p-value<0.01
Step 7. Compute the statistic.: __________________
Evidence: n=9,
x  4.2
t
s  1.5
x   4.2  3.4 0.8


 1.6  2.896, do not reject H 0
s
1.5
0.5
n
9
Compute the p-value:
 2nd DIST 6 tcdf( 1.6, EE5, 8)
ENTER
p-value is approximately 0.074 >0.01
Step 8. Conclusion: do not Reject H0. We have statistical evidence at a 0.01 level of
significance to believe that the evidence is in line with the manufacturer's claim that the average
nicotine content does not exceed 3.4 milligrams
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3. It is believed that less than 60% of the residents in a certain area favor an annexation suit by a
neighboring city. What conclusion would you draw if only 110 in a sample of 200 voters favor the
suit? Use a 0.04 level of significance.
Step 1. State Ho: _ p = 0.60________________________
State H1: _ p < 0.60 ____________________________
Step 2. Type of test; _left- tailed test_________
Step 3. Level of significance: _
=0.04__________
Step 4. critcal value of the statistic: _ z = - 1.75 ____
 2nd DISTR 3 invNorm(0.04) ENTER answer: -1.750686071
Step 5. Diagram
0.04
p  0.60
z=-1.75
Step 6. Decision rule: _Reject H0 if z computed from evidence is <--1,75
or p-value<0.04
Step 7. Compute the statistic.: ____-1.44 ______________
Evidence: n=200,
x 110
x  110 pˆ  
 0.55
n 200
z
pˆ  p
pq
n

0.55  0.60
(0.60 )(. 40 )
200

 0.05
 1.44
0.0346
Compute the p-value

2nd DIST 2 normalcdf(-EE5, -1.44) ENTER
p-value = 0.0749 > 0.04
Step 8. Conclusion: do not Reject H0. We have statistical evidence at a 0.04 level of
significance to believe that the proportion of the residents in a certain area who favor an
annexation suit by a neighboring city is not less than 60%.
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4. A food company is planning to market a new type of product. However, before marketing this
product, the company wants to find what percentage of the people like it.
The company’s management has decided that it will market the new product only if at least 35%
of the people like it. The company’s research department selected a random sample of 400
persons and asked them to taste this yogurt. Of these 400 persons, 112 said they liked it. Testing
at the 2.5 % significance level , can you conclude that the company should market this new
product?
Show all the required steps and give a precise conclusion.
Step 1. State Ho: __ p≥ 0.35 _______________________
State H1: ___p
<0.35 ___________________________
Step 2. Type of test; ___ left - TAILED TEST__________
Step 3. Level of significance: __
 = 0.025__
Step 4. critcal value of the statistic: Z=-1.96
 2nd DIST 3 invNormal(0.025) ENTER answer: Z=-1.959963986
Step 5. Diagram
0.025
z=-1.96
p  0.35
Step 6. Decision rule: REJECT
or p-value < 0.025
HO IF Z COMPUTED FROM EVIDENCE IS <-1.96
Step 7. Compute the statistic: __________________
112
Evidence: n=400, x-112, pˆ 
 0.28
400
z
pˆ  p
pq
n

0.28  0.35
(0.35 (0.65 )
400

0.07
 2.94
0.0238
Compute the p-value
 2nd DIST 2 normalcdf (-EE5, -2.94) ENTER answer: 0.0016411295
p-value= P(z<-2.94) = 0.0016 <0.025
Step 8. Conclusion:
Reject Ho. We have statistical evidence at a 2.5% level of significance to believe that the
company should not market the yogurt because the evidence doesn’t show that at least 35% of
the people like it.
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5. According to a Census survey , American workers aged 30 to 45 spent an average of 22
minutes traveling to. A recently taken random sample of 20 American workers in this age group
yielded a mean commuting time to work equal to 25 minutes with a standard deviation of 8
minutes. Assume that the times to commute to work for all such workers are normally distributed.
Using a 5 % significance level, can you conclude that the current mean travel times to work for all
U.S. workers ages 30 to 45 is different than 22 minutes?
Step 1. State Ho: ___ =22______________________
State H1: _____≠22_________________________
Step 2. Type of test; _two tailed test_________
Step 3. Level of significance: _
=0.05 __
Step 4. critcal value of the statistic: t0.025, 19  2.093

2nd DIST 4 invT(0.025, 19)
answer=-2.093024022
Step 5. Diagram
0.025
t=-2.093
0.025
t=2.093
  22
Step 6. Decision rule: _Reject H0 if t computed from evidence is <-2.093 or >2.093
or if p-value <0.05
Step 7. Compute the statistic.: __________________
Evidence: n=20,
x  25 s  8
t
x   25  22
3


 1.68
s
8
1.789
n
20
Compute the p-value
 2 2nd DISTR 6 tcdf(1.68, EE3, 19) ENTER
p-value=2P(t>1.68)=2(0.5466)=0.109 >0.05
answer: 0.1093254795
Step 8. Conclusion: DO NOT REJECT HO , we have statistical evidence at a
5 % level OF SIGNIFICANCE to believe that the current mean travel times to work for all U.S.
workers ages 26 to 44 is 22 minutes.
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