Download Chapter 6 Section 2

Chapter 6 Section 2
Homework A
6.51 What's wrong? Here are several situations where there is an incorrect application of the ideas presented in
this section. Write a short paragraph explaining what is wrong in each situation and why it is wrong.
(a) A significance test rejected the null hypothesis that the sample mean is equal to 1500.
The null hypothesis is about the population mean not the sample mean. The sample mean only provides
evidence about what is the value of the population mean.
(b) A change is made that should improve student satisfaction with the way grades are processed- The null
hypothesis, that there is an improvement, is tested versus the alternative, that there is no change.
This is backwards. The null hypothesis has to be the status quo- no change.
(c) A study summary says that the results are statistically significant and the P-value is 0.99.
A statistically significant result statest that the p-value is very small, which means our result is rare when we assume the
null hypothesis is true. A result with a p-value of 0.99 would not be statistically significant.
6.53 More on determining hypotheses. State the null hypothesis H0 and the alternative hypothesis Ha in each
case. Be sure to identify the parameters that you use to state the hypotheses.
(a) A university gives credit in French language courses to students who pass a placement test. The language
department wants to know if students who get credit in this way differ in their understanding of spoken French from
students who actually take the French courses. Experience has shown that the mean score of students in the courses
on a standard listening test is 26. The language department gives the same listening test to a sample of 35 students
who passed the credit examination to see if their performance is different.
The null hypothesis states that the students who got credit for French via the placement test have a mean of
26; the same value as the students who take the course.
Why?
One could say that the students that get credit via taking the test should be as good as the ones that take the course.
Another issue is that we do not have a model for the students that take the placement test. If not  = 26 what would it
be? Why not assume it is the same as the other students that we have information on and see if we can spot a difference.
The alternative hypothesis says that the mean score is something else other than 26,   26.
examination to see if their performance is different.
Why? Because of the statement above says different.
(b) Experiments on learning in animals sometimes measure how long it takes a mouse to find its way through a
maze. The mean time is 20 seconds for one particular maze. A researcher thinks that playing rap music will cause
the mice to complete the maze faster. She measures how long each of 12 mice takes with the rap music as a
stimulus.
H0:  = 20 seconds Why? We want to show that the mice will run the maze faster with the music playing,
but we cannot assume what we want to provide evidence of. But ask yourself, if not 20 seconds for the null
what should it be?
Ha:  < 20 seconds
(c) The average square footage of one-bedroom apartments in a new student-housing development is advertised to
be 460 square feet. A student group thinks that the apartments are smaller than
6.55 Translating research questions into hypotheses.
Translate each of the following research questions into appropriate Ho and Ha.
(a) Census Bureau data show that the mean household income in the area served by a shopping mall is $62,500 per
year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall
shoppers is higher than that of the general population.
(b) Last year, your company's service technicians took an average of 2.6 hours to respond to trouble calls from
business customers who had purchased service contracts. Do this year's data show a different average response
time?
It would be wise to use last years average as the starting point of the investigation.
H0:  = 2.6 hours.
The statement “show a different average” is interpreted to the alternative,
Ha:  ≠ 2.6 hours.
6.57 More on computing the P-value. A test of the null hypothesis Ho:  = 0 gives test statistic z = -1.73.
For the following scenarios below, draw a picture of the null situation using the z-distribution (note z = 0),
with the z-statistic labeled and the area corresponding to the p-value shaded. Use function notation as
shown in-class to communicate your answers.
For example the labeled graph for (a) should be
(a) What is the P-value if the alternative is Ha:  > 0?
P(Z > - 1.73) = 0.9581
Entered: =1 – normsdist(-1.73) in Excel.
(b) What is the P-value if the alternative is Ha:  < ?
P(Z < - 1.73) = 0.0419
Entered: =normsdist(-1.73) in Excel.
6.61 Purity of a catalyst. A new supplier offers a good price on a catalyst used in your production process. You
compare the purity of this catalyst with that of the catalyst offered by your current supplier. The P-value for
a test of "no difference" is 0.27. Can you be confident that the purity of the new product is the same as the
purity of the product that you have been using? Discuss.
When this type of test is run, what is the null hypothesis? The question itself says a test of no
difference, that is, the null hypothesis is assumed that the purity of the catalysts is exactly the same.
Now the actual sample produced a difference that was not zero, because if it was, then we would have
a z-score of 0, and when we calculate P(Z > 0) = 0.5, the p-value; this assumes the distribution of
differences is normally distributed. Instead what we have is that the purity of one in the sample was
better than the other, but the calculation
shows that we would observe a difference as
large as the one we observed or an even
larger difference 27% of the time, if we
assumed there was no difference. A 27% Pvalue is not rare enough to say that there is






 Z  evidence to suggest that the two purity levels
are different. So at the moment the evidence
that there is a difference is very very weak.
1. Do you know how to read a p-value? - A drug is being tested to see if it can outperform a placebo. The
null hypothesis will be that the drug is no better than a placebo, while the alternative will be the drug is
outperforming the placebo. Which p-value gives a stronger evidence against the null hypothesis: p-value =
0.65, p-value = 0.45, p-value = 0.26, or p-value = 0.06?
The p-value 0.06 provides the most evidence against the null hypothesis.
The lower the p-value the more evidence we have against the null hypothesis. Why? The p-value
signifies the likelihood that the number we have observed or an even more extreme value (further
away from the expected parameter, in 6.2 the parameter is the population mean ) will show up if the
null value is correct. Thus if your value belongs to a group of rare numbers you wonder why a rare
value appeared; after all a rare number is a value that hardly ever occurs when the null hypothesis is
true. So we start wondering if the reason it showed up is because the null hypothesis is not correct.
2. Reading the sample statistic. – A test is conducted in which the random variable X is thought to have a
normal distribution with  = 15.4 and  = 2.3. A test is conducted to see if there is any evidence that
Ha:  < 15.4, where Ho:  = 15.4. A random sample of 15 is gathered
(a) Which of the three sample means provide more evidence against the null? x = 11.5, x = 12.7 or x = 15.6?
The value x = 11.5 provides the most evidence against the null since it is further away from our expected
value of 15.4.
(b) Lets say that the sample average is 13.5. What is the p-value?
Test statistic: z =
13.5  15.4
2.3
15
= -3.19
x = 13.5
P  X < 13.5  = P(Z < - 3.19)
X
μ x  15.4
= 0.000711
The p-value is 0.000711
6.69 Attitudes toward school. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that
measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200.
The mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher who
suspects that older students have better attitudes toward school gives the SSHA to 25 students who are at
least 30 years of age. Their mean score is x = 132.2.
(a) Assuming that  = 30 for the population of older students, carry out a test of
Ho:  = 
Ha:  > 
Report the P-value of your test, and state your conclusion
132.2  115
clearly. Test statistic: z =
30
25
= 2.86
P  X > 132.2  = P(Z > 2.86)
= 0.0021
The p-value is 0.0021 which indicates we have really good
evidence that the older students have on average a higher
attitude score. An average as high as 132.2 from a sample of
25, or even a higher average if we assume the true average is
115 would occur 21 times out of 10,000 atttempts in the long
run; we have just witnessed a very rare event if we assume
that the null hypothesis is correct.
(b) Your test in (a) required two important assumptions in addition to the assumption that the value of  is known.
What are they? Which of these assumptions is most important to the validity of your conclusion in (a)?
We had to assume that the sample is indeed representatitve (unbiased sample) of the population of older
students. We also assumed the distribution of the sample averages is normally distributed or nearly so.
3. Quality Control – Between 1995 and 2002 Southwest Airlines had customers get a boarding pass upon arrival,
then to board the plane the passengers would line up line up single file. The passengers were broken up into three
groups, called A, B and C. When the people assigned A would be called to board, the people in A group would get
to enter the plane depending on where they are on the line. The boarding procedure would take 18 minutes on
average with a standard deviation of 3.5 minutes, to load 148 passengers.
A new procedure was implemented in which the boarding pass gave you a position in the line along with a letter
designation, furthermore, poles were installed with number designations (1 – 5, 5 – 10, and so on) so that
passengers knew where to line up. The idea was to decrease boarding time, and also to decrease confusion and
animosity among passengers.
a. Let us say that this procedure was implemented at two airports. A sample of 120 flights yield a sample mean
boarding time of 19.3 minutes. Does this sample provide any evidence that the boarding time has decreased?
Explain why or why not.
Under the previous procedure the mean boarding time was 18 minutes. Our new procedure yielded a
sample mean of 19.3 minutes, thus the sample mean shows no improvement in boarding time.
b. Let us say that the procedure is implemented at three airports. A sample of 50 flights is gathered, and the
sample average is 17.1, calculate the amount of evidence that this sample average provides against the null
hypothesis.
Ho:  = 
Ha:  < 18.0
Test statistic: z =
17.1  18.0
3.5
50
= -1.818
P  X < 17.1 = P(Z < - 1.818)
The calculated amount of evidence against the null
hypothesis is 0.03453, which is the p-value. This means that if 
= 18.0 minutes with  = 3.5 minutes, we would see a sample
average of 50 values less than or equal to 17.1 minutes around
3.453% of the time. Thus, 17.1 minutes or less is not a common
sample average when  = 18.0 minutes.
= 0.03453
4. Understanding the P-value. - Choose the correct value. The p-value is the probability of observing a
value as large as the one observed or even larger assuming that
a. the null value 0 is correct
b. the null value 0 is incorrect.
The p-value is always calculated assuming that the null value is correct.