Download Conduct a hypothesis test for each problem, using the traditional

Conduct a hypothesis test for each problem, using the traditional method. Show the 5
steps and all work for each hypothesis test
Step 1: State the hypotheses and identify the claim
Step 2; Find the critical values f
Step 3 Compute the test value
Step 4: Make the decision to reject or not reject the null hypothesis.
Step 5: Summarize the results.
1. A random sample of 56 departing flights at an airport over a 3-month period had a
mean wait of 14.2 minutes between boarding and takeoff, with a standard deviation of
4.53 minutes. At the same airport, a random sample of 81 incoming flights over the same
3-month period had a mean wait of 17.5 minutes between the time that the plane arrived
at the gate and the time that the baggage reached the baggage claim area, with a standard
deviation of 9.87 minutes. At the 0.05 level of significance, test the claim that at this
airport the mean wait for takeoff is less than the mean wait for baggage.
Solution: Let 1 be the mean wait time between boarding and takeoff, and 2 be the
mean wait time between the time that the plane arrived at the gate and the time that the
baggage reached.
Step 1: The hypothesis testing problem is H0 : 1 = 2 against Ha : 1 < 2.
Here the alternative hypothesis is the claim.
Step 2: The level of significance is  = 0.05. The critical value of t at 119 degrees of
freedom and for 0.05 level of significance is t-critical = -1.6578.
Step 3: The test statistics is
𝑡=
𝑥̅1 − 𝑥̅2
𝑠2
√ 1
𝑠22
𝑛1 + 𝑛2
=
14.2 − 17.5
2
2
√4.53 + 9.87
81
56
= −2.6344
Step 4: As the test statistics < t-critical, we reject the null hypothesis.
Step 5: The claims that at the airport the mean wait for takeoff is less than the mean wait
for baggage.
2. A math instructor has written a new placement exam. He is concerned that the exam
may be too long. The instructor claims that the average length of time required to finish
the exam for all students who take it is at most 50 minutes. The exam is given to 20
randomly selected students, and their average length of time required to finish the exam
was 54 minutes, with a standard deviation of 5.99 minutes. Test the instructor's claim at
the 0.05 level of significance.
Solution:
Let  be the average length of the time required to finish the exam for all students who
take it.
Step 1: The hypothesis testing problem is H0 :   50 min against Ha :  > 50 min.
Here the null hypothesis is the claim.
Step 2: The level of significance is  = 0.05. The critical value of t at 19 degrees of
freedom and for 0.05 level of significance is t-critical = 1.7291.
Step 3: The test statistics is
𝑡=
𝑥̅ − 𝜇
𝑠⁄√𝑛
=
54 − 50
5.99⁄√20
= 2.9864
Step 4: As the test statistics > t-critical, we reject the null hypothesis.
Step 5: The claims that the average length of time required to finish the exam for all
students who take it is at most 50 minutes is rejected. So the average length of time
required to finish the exam for all students who take it is more than 50 minutes
3. A physician claims that 13.6% of all pregnant females smoke during their pregnancy.
A random sample of 400 pregnant women revealed that 60 of them are smoking while
pregnant. Test the physician's claim at the 0.05 level of significance.
Solution:
Let p be the proportion of pregnant females who smoke during their pregnancy.
Step 1: The claim that the 13.6 % of the females smoke during their pregnancy. It can be
expressed as p = 0.136. So the hypothesis testing problem can be written as
H0 : p = 0.136 against Ha : p  0.136 (Two-tailed). Here null hypothesis is the claim.
Step 2: The level of significance is  = 0.05. So the critical values of Z are -1.96 and
1.96. (Since the alternative hypothesis is two-tailed)
𝑋
60
Step 3: 𝑝̂ = 𝑛 = 400 = 0.15
The test statistics is
𝑍=
𝑝̂ − 𝑝
√𝑝̂ (1 − 𝑝̂ )
𝑛
=
0.15 − 0.136
√(0.15 ∗ 0.85)
400
= 0.8168
Z = 0.8168.
Step 4. Since the Z-test statistics falls outside the critical region, we do not reject the null
hypothesis.
Step 5: A physicians claim is accepted at 0.05 level of significance. That is 13.6% of all
pregnant females smoke during their pregnancy.
4. A random telephone survey asked 208 drivers if they run red lights. Of the 48 drivers
who were 35 years old or younger, 32 admitted that they run red lights. Eighty-six of the
160 drivers that were older than 35 years old admitted to running red lights. At the 0.01
level of significance, test the claim that the percent of drivers who are 35 years old or
younger and who run red lights is higher than the percent of drivers who are older than 35
years old and who run red lights.
Solution:
Let p1 be the proportion of drivers who are 35 years old or younger and who run red
lights and p2 be the proportion of drivers who are older than 35 years old and who run
red lights
Step 1: The claim that the percent of drivers who are 35 years old or younger and who
run red lights is higher than the percent of drivers who are older than 35 years old and
who run red lights p1 > p2.
Thus the hypothesis testing problem is
H0 : p1  p2 against H1 : p1 > p2 (original claim)
Step 2: The significance level is 0.01. So the critical value of Z at 0.01 level of
significance is Z-critical = 2.3263.
Step 3: We will use the normal distribution as an approximation to the binomial
distribution. We estimate the common value of p1 and p2 with the pooled sample
estimate calculated as
𝑝̅ =
𝑥1 + 𝑥2
32 + 86
=
= 0.5673
𝑛1 + 𝑛2 48 + 160
𝑞̅ = 1 − 𝑝̅ = 1 − 0.5673 = 0.4327
Step 6 : The test statistics is
𝑍=
(𝑝̂1 − 𝑝̂2 ) − (𝑝1 − 𝑝2 )
𝑝̅ 𝑞̅ 𝑝̅ 𝑞̅
𝑛1 + 𝑛2
√
31
16
(2823 −
)−0
7765
=
√(0.004439 ∗ 0.995561) + (0.004439 ∗ 0.995561)
2823
7765
Z = 6.105785
This is a right-tailed test, so the P-value is the area to the right of the test statistic
(Z=6.105785). The area to the right of the test statistic is 0.00000, so the P-value is
0.00000.
Step 7: Because the P-value of 0.00000 is less than the significance level of  = 0.05, we
reject the null hypothesis. That is the fatality rate is higher for those not wearing seat
belts. The use of seat belts appears to be effective in saving lives.
5. Test the claim that the mean time Americans spend watching TV per day is 4.4 hours
at the 0.10 level of significance. A random sample of 80 people had a mean time of
2.225 hours viewing per day and a standard deviation of 1.088 hours.
Solution:
Let  be the mean time Americans spend watching TV per day.
Step 1: The claim is the mean time Americans spend watching TV per day is 4.4 hours
Hence the hypothesis testing problem is H0:  = 4.4 hours against Ha:   4.4 hours.
Here the null hypothesis is the claim.
Step 2: The level of significance is  = 0.10. The critical values of t at 79 degrees of
freedom and for 0.10 level of significance are t-critical -1.644 and 1.644 (Since the
alternative hypothesis is two-tailed.
Step 3: The test statistics is
𝑡=
𝑥̅ − 𝜇
𝑠⁄√𝑛
=
2.225 − 4.4
1.088⁄√80
= −17.8803
Step 4: As the test statistics < -1.644, we reject the null hypothesis.
Step 5: The claims that the mean time Americans spend watching TV per day is 4.4
hours is rejected. So the time Americans spend watching TV per day is not equal to 4.4
hours.
6. Tests in a professor's past statistics classes have scores with a standard deviation equal
to 14.1. One of her current classes now has 27 test scores with a standard deviation of
9.3. Use a 0.01 level of significance to test the claim that the standard deviation of this
current class is less than 14.1.
Solution:
Let  be the standard deviation of scores of past statistics classes.
Step 1: The claim is that the standard deviation of current class scores is less than 14.1.
So the hypothesis testing problem can be written as H0 :  = 14.1 against Ha:  < 14.1
(Claim).
Step 2: The level of significance is  = 0.01. The critical value of the 2 at n – 1 (= 26)
degrees of freedom and at 0.01 level of significance is
2
𝜒28,0.01
= 12.1981
Step 3: The test statistics is
𝜒2 =
𝑛𝑠 2 27 ∗ 9.3
=
= 159.485
𝜎2
14.1
Step 4: As the test statistics falls in the critical region (that is 2 < 2 –critical) we do not
reject the null hypothesis.
Step 5: Hence the claim that the standard deviation of the scores of current class scores is
less than 14.1 is rejected.