Solutions to Worksheet on Hypothesis Tests

1. A production line produces rulers that are supposed to be 12 inches long. A sample of 49 of the rulershad a mean of 12.1 and a standard deviation of .5 inches. The quality control specialist responsible forthe production line decides to do a hypothesis test at the 90 percent significance level to determinewhether the production line is really producing rulers that are 12 inches long or not.a. What is the null hypothesis?

Solution:H0 : µ = 12

b. What is the alternative hypothesis?Solution:H1 : µ ≠ 12

c. What is the value of the test statistic?Solution:

z =x − µ0s

n=

12.1 − 12.5

49= 1.4

d. What is the rejection region (with its numerical value)?Solution:z > zα 2

= z.05 = 1.645e. What conclusion do you draw?

Solution:Do not reject H0 since | 1.4| is not greater than 1.645.

f. What does this mean in terms of the problem situation?Solution:We cannot conclude that the mean length of the population of rulers from the production line is notequal to 12. We should continue with our working assumption that it is equal to 12.

2. A sample of 20 OU freshmen had a mean GPA of 2.8 over all their courses taken in their first semester atOU. This had a variance of .25. Perform a hypothesis test at the 95 percent level to determine if the firstsemester GPA of all OU freshmen is less than a B (3.0).a. What is the null hypothesis?

Solution:H0 : µ = 3.0

b. What is the alternative hypothesis?Solution:H1 : µ < 3.0 (one-sided test, "less than"

c. What is the value of the test statistic?Solution:

t =x − µ 0s

n=

2.8 − 3.0.5

20≈ −1.79 (small sample)

d. What is the rejection region (with its numerical value)?Solution:t < −t n−1,α = −t19,.05 = −1.729

e. What conclusion do you draw?Solution:Reject H0 since -1.79 < -1.729.

f. What does this mean in terms of the problem situation?Solution:We must reject our working hypothesis that the mean GPA is 3.0 or more and conclude that it isless than 3.0.

3. A researcher decides to look at the variance of the production line in Problem 1 She decides to do ahypothesis test at the 90 percent significance level to determine if the variance is actually less than .25.a. What is the null hypothesis?

Solution:H0 : σ 2 = .25

b. What is the alternative hypothesis?Solution:H1 : σ 2 < .25

c. What is the value of the test statistic?Solution:

χ 2 =n − 1( )s2

σ02 =

49 − 1( ) .5( )2

.25= 48

d. What is the rejection region (with its numerical value)?Solution:χ 2 < χ n−1,1 −α

2 = χ 48,.902 ≈ 29.050

e. What conclusion do you draw?Solution:We cannot reject H0 since 48 is not less than 29.05

f. What does this mean in terms of the problem situation?Solution:We must continue with our working assumption that the true variance is .25 or more and cannotassume that it is less than .25.

4. A math teacher for Math 4753 wants to determine if a new book proposed for the course is better thanthe old book that was used before. The math teacher decides that the two classes might have differentmath ability coming into the course. He gives a test at the beginning of the semester to measure themath ability the students have when they come into the course. He uses this test to create a matched pairdesign for 30 students from each class. The scores he gets are as follows:

Student Pair New Book Class Old Book Class1 96 972 95 943 94 944 92 955 92 906 90 917 90 908 89 859 88 8710 88 9211 86 8512 86 8113 84 8914 84 8315 84 8516 83 8017 83 7218 81 8919 79 7620 78 7821 77 7522 77 7623 72 7424 70 8125 68 7126 67 6627 65 6528 64 6829 62 6530 62 58

Perform a hypothesis test at the 95 percent level for the matched pairs design in Problem 8 to see if thescores for the new book are higher than those for the old book.a. What is the null hypothesis?

Solution:H0 : µ D = µ1 −µ 2 = 0 ( µ1 is the mean for the new book)

b. What is the alternative hypothesis?Solution:H1 : µ D = µ1 −µ 2> 0

c. What is the value of the test statistic?Solution:First we need to find the differences, their mean and their variance:

Student Pair New Book Class Old Book Class Differences1 96 97 -12 95 94 13 94 94 04 92 95 -35 92 90 26 90 91 -17 90 90 08 89 85 49 88 87 1

10 88 92 -411 86 85 112 86 81 513 84 89 -514 84 83 115 84 85 -116 83 80 317 83 72 1118 81 89 -819 79 76 320 78 78 021 77 75 222 77 76 123 72 74 -224 70 81 -1125 68 71 -326 67 66 127 65 65 028 64 68 -429 62 65 -330 62 58 4

d = -0.2sd

2 = 16.7172414

The value of the test statistic is then t =d − δ 0sd

n=

−.2 − 016.1724

30≈

−.2.7342

≈ −0.273

d. What is the rejection region (with its numerical value)?Solution:t > tn −1,α = t29,.05 = 1.699

e. What conclusion do you draw?Solution:We cannot reject H0 since -0.273 is not greater than 1.699.

f. What does this mean in terms of the problem situation?Solution:We must continue to use the working assumption that the means are not different and cannotconclude that the means for the new book are higher than for the old book.

5. Five engineering students and 5 computer science students are arranged in a matched pair design on thebasis of their GPA prior to the last semester of their senior year. Their GPAs for their last semester areas follows:

Pair Number Engineering Major CS Major1 3.89 3752 3.61 3.873 2.80 3.024 2.75 2.925 3.41 1.28

Perform a 90 percent hypothesis test to determine whether the means for CS majors are less than themeans for Engineering majors.a. What is the null hypothesis?

Solution:H0 : µ D = µ1 −µ 2 = 0 (This is Engineers minus CS majors)

b. What is the alternative hypothesis?Solution:H1 : µ D = µ1 −µ 2> 0 (Since this is Engineering majors minus CS majors and we are interestedin whether the means for CS majors are less, we can answer the question by seeing if theEngineering majors are more and that's what we are doing since we have already found thedifferences as Engineering majors minus CS majors and not the other way around.)

c. What is the value of the test statistic?Solution:First we need to find the differences, their mean and their variance:

Pair Number EngineeringMajor

CS Major Differences

1 3.89 3.75 0.142 3.61 3.87 -0.263 2.8 3.02 -0.224 2.75 2.92 -0.175 3.41 1.28 2.13

d = 0.324sd

2 = 1.04413

So t =d − δ 0sd

n=

.324 − 01.04413

5≈

.324.4570

≈ 0.709

d. What is the rejection region (with its numerical value)?Solution:t > tn −1,α = t4, .1 = 1.533

e. What conclusion do you draw?Solution:We cannot reject H0 since .709 is not greater than 1.533.

f. What does this mean in terms of the problem situation?Solution:We must continue to assume that there is no difference between the grades for the two majors andcannot conclude that CS majors have lower grades.

6. The teacher in Problem 5 is still not sure about the differences in Problem 5 between CS and Engineeringmajors. He takes independent samples of size 10 from each major at random. He does not use theirprevious GPA to make matched pairs. The 10 Engineering majors have a mean last-semester GPA of2.89 with a variance of .5. The 10 CS majors have a mean last semester GPA of 2.66 with a variance of.3. He decides to do a hypothesis test to see if the GPAs are different for the two populations. He doesthe test at the 90 percent level.a. What is the null hypothesis?

Solution:H0 : µ1 − µ 2= 0

b. What is the alternative hypothesis?Solution:H1 : µ1 − µ 2≠ 0

c. What is the value of the test statistic?Solution:

t =x − y − δ 0

s 1n1

+1n2

=2.89 − 2.66 − 0

.17( ) 110

+110

≈0.23

0.1844≈ 1.247

where s 2 =n1 − 1( )s1

2 + n2 − 1( )s22

n1 + n2 − 2=

10 − 1( ) .5( )2 + 10 − 1( ) .3( )2

10 + 10 − 2=

3.0618

= 0.17

d. What is the rejection region (with its numerical value)?Solution:t > tn1 +n2−2 ,α 2

= t18,.05 = 1.734e. What conclusion do you draw?

Solution:We cannot reject H0 because 1.247 is not greater than 1.734.

f. What does this mean in terms of the problem situation?Solution:We cannot conclude that the two means are different since we cannot reject the null hypothesis.

7. Suppose in the situation in Problem 6, the teacher had the GPAs for a sample of 10 Engineering majorsbut only 8 CS majors but otherwise the data were the same. Suppose he decided to do a 90 percenthypothesis test for the difference in GPAs for the two populations of majors.a. What is the null hypothesis?

SolutionH0 : µ1 − µ 2= 0

b. What is the alternative hypothesis?SolutionH1 : µ1 − µ 2≠ 0

c. What is the value of the test statistic?Solution

t =x − y − δ0

s12

n1+

s12

n2

=2.89 − 2.66 − 0

.5( )2

10+

.3( )2

8

≈0.23

0.1904≈ 1.208

d. What is the rejection region (with its numerical value)?Solutiont > tν,α 2

= t15,.05 = 1.753

where ν =

s12

n1+

s22

n2

2

s12

n1

2

n1 − 1+

s22

n2

2

n2 − 1

=

.5( )2

10+

.3( )2

8

2

.5( )2

10

2

10 − 1+

.3( )2

8

2

8 − 1

=.03625( )2

8.7525 × 10 −5 ≈ 15.01 ≈ 15

e. What conclusion do you draw?SolutionWe cannot reject H0 because 1.208 is not greater than 1.753.

f. What does this mean in terms of the problem situation?SolutionWe cannot conclude that the means are different and must continue to assume they are equal.

8. The math teacher for Math 4753 wants to determine if the new book proposed for the course is betterthan the old book that was used before. He teaches two classes that are each a random sample of 42people. The class using the new book got an average of 88 on the final exam given by the math teacherand had a standard deviation of 3 points. The class using the old book had an average of 90 with astandard deviation of 2 points. Suppose the teacher decides to do a hypothesis test at the 98 percentlevel to see if the variances for Engineering majors is lower than the variances for CS majors.a. What is the null hypothesis?

Solution:H0 : σ1

2 ≤ σ22

b. What is the alternative hypothesis?Solution:H1 : σ1

1 > σ 22

c. What is the value of the test statistic?Solution:

F =s1

2

s22 =

3( ) 2

2( )2 = 2.25

d. What is the rejection region (with its numerical value)?Solution:F > fn1 −1, n2 −1,α = f41,41,.01 ≈ 2.11 using the values for 40,40 rather than for 41,41.

e. What conclusion do you draw?Solution:We can reject H0 since 2.25 is greater than 2.11.

f. What does this mean in terms of the problem situation?Solution:We conclude that the variance for the new books is higher than the variance for the old books.

9. A survey of 4000 people in the US finds that 2856 of them believe that daily weather reports are totallyuseless because meteorology is not really a science. Given this data perform a 95 percent hypothesis testto see if more than half of the people in the US believe that weather reports are useless.a. What is the null hypothesis?

Solution:H0 : p ≤ .5

b. What is the alternative hypothesis?Solution:H1 : p > .5

c. What is the value of the test statistic?Solution:

z =ˆ p − p0

p0q0

n

=.714 − .5

.5 × .54000

≈.214.0079

≈ 27.09

d. What is the rejection region (with its numerical value)?Solution:z > zα = z.05 ≈ 1.645

e. What conclusion do you draw?Solution:We can reject H0 since 27.09 > 1.645.

f. What does this mean in terms of the problem situation?Solution:We can conclude that it is likely (at the 95% confidence level) that more than half of the people inthe US believe that weather reports are useless.

10. A survey of 200 regular viewers of Channel 5 in Oklahoma City show that 68 believe that GaryEngland is God. A survey of 100 regular viewers of Channel 9 in Oklahoma City show that 68 of themalso believe that Gary England is God. Given these data perform a 90 percent hypothesis to determine ifthe proportion of Channel 9 viewers who believe that Gary England is God is greater than theproportion of Channel 5 viewers who believe it.a. What is the null hypothesis?

Solution:H0 : p2 − p1 ≤ δ 0 = 0 (Population 2 are the regular Channel 9 viewers)

b. What is the alternative hypothesis?Solution:H1 : p2 − p1 > δ 0 = 0

c. What is the value of the test statistic?Solution:

z =ˆ p 2 − ˆ p 1

ˆ p ̂ q 1n1

+1n2

=.68 − .34

.4533 × .5467 1200

+1

100

≈.34

.06097≈ 5.58

where ˆ p =x + y

n1 + n2=

68 + 68200 + 100

≈ 0.4533 and we use this form of the test statistic because

δ0 = 0 (see Equation 9.12).d. What is the rejection region (with its numerical value)?

Solution:z > zα = z.1 ≈ 1.285

e. What conclusion do you draw?Solution:Reject H0 because 5.58 > 1.285.

f. What does this mean in terms of the problem situation?Solution:This means that we can conclude that H1 is true and that p2 − p1 > 0 or, in other words, p2 > p1(a greater proportion of regular Channel 9 viewers believe that Gary England is God than theproportion of regular Channel 5 viewers that believe that Gary England is God. [Oh, please, Gary,forgive the Channel 5 viewers for they know not what they do]).