Hypothesis Testing Using a Single Sample

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Chapter 10

• Hypothesis Testing Using a Single Sample

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Chapter 10 Hypothesis Testing

• An Example A report released by the National
  Association of Colleges and Employers stated that
  the average salary for students graduating in
  2006 with a degree in accounting was 45,656.
• Suppose that you are interested in
  investigating whether the mean starting salary
  for students graduating with an accounting degree
  from your university this year is gt 45,656, a
  sample of n 40 accounting graduates from the
  university was selected, and it was found that
  the mean starting salary is 45,958 with a
  standard deviation of 1214.
• Is µ gt 45,656 a reasonable conclusion?

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10.1 Hypotheses and Test Procedures

• A test of hypotheses or test procedures is a
  method for using sample data to decide between
  two competing claims (hypotheses) about a
  population characteristic.
• One hypothesis µ 45,656, and the other µ?
  45,656
• One hypothesis µ 45,656, and the other may be
  µ gt 45,656.
• We initially assume that a particular hypothesis
  (the null hypothesis) is the correct one. Then we
  consider the sample data, and if there is
  convincing evidence against the null hypothesis,
  we reject the null hypothesis in favor of the
  competing hypothesis.

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Null and Alternative Hypotheses

• The null hypothesis, denoted by H0, is a claim
  about a population characteristic that is
  initially assumed to be true.
• The alternative hypothesis, denoted by Ha, is the
  competing claim.
• The two possible conclusions are
• reject H0 if sample evidence strongly suggests
  that H0 is false or
• fail to reject H0 if the sample does not contain
  such evidence.
• Similar to US judicial system
• H0 The defendant is innocent versus
• Ha The defendant is guilty
• Rejecting H0 means that the jury
  find the defendant guilty, while failing to
  reject H0 means that the jury find the defendant
  not guilty.

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• The form of a null hypothesis is
• H0 population characteristic hypothesized
  value
• where the hypothesized value is a specific
  number determined by the problem context.
• The alternative hypothesis has one of the
  following three forms
• Ha population characteristic gt hypothesized
  value
• Ha population characteristic lt hypothesized
  value
• Ha population characteristic ? hypothesized
  value

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• Example Tennis Ball Diameter
• Because of variation in the manufacturing
  process, tennis balls produced by a particular
  machine do not have identical diameters. Suppose
  the machine was originally calibrated to achieve
  the design specification of 3 in diameter.
  However, the manufacturer is now concerned that
  the diameters no longer conform to this
  specification. What is the sensible choice of
  hypotheses?

H0 µ 3 (the specification is being met, so
recalibration is unnecessary Ha µ ? 3 (the
specification is not being met, so recalibration
is necessary)
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• Example Light Bulb Lifetimes
• Kmart brand 60-W light bulbs state on the
  package Ave. Life 1000 Hr. People who purchase
  this brand would be unhappy if the actual life
  time is less than the advertised 1000 hours.
  Suppose a sample of Kmart light bulbs is selected
  and the lifetime for each bulb in the sample is
  recorded. Choose the null and alternative
  hypotheses for the test.
• People who purchase this brand would be unhappy
  if µ is actually less than 1000 hours. A sample
  of Kmart light bulbs is selected and the lifetime
  for each bulb in the sample is recorded. The
  sample results can then be used to test the
  hypothesis µ 1000 hours against the hypothesis
  µ lt 1000.

H0 µ 1000 Ha µ lt 1000
H0 will be rejected only if sample
evidence strongly suggests that µ 1000 is not
plausible.
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10.2 Errors in Hypothesis Testing

• Just like a jury may reach the wrong
  verdict in a trial, there is some chance that
  using a test procedure with sample data may lead
  us to wrong conclusion about the population
  characteristics.
• Two types of error that might be made using a
  hypothesis test

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• Example On-Time Arrival The US Department of
  Transportation reported that during a recent
  period, 78.6 of all domestic passenger flights
  arrived on time. Suppose that an airline decides
  to offer its employees a bonus, in an upcoming
  month, if the airlines proportion of on-time
  flights exceeds the overall industry rate of
  0.786. Let p be the true proportion of the
  airlines flights that are on time during the
  month. Set a hypothesis test for p and discuss
  the types of error.
• Solution H0 p .786 Ha p gt .786
• Type I error (Reject a true H0) The Airlines
  reward its employees when in fact their true
  proportion of on-time flights did not exceed
  78.6
• Type II error (Fail to reject a false H0) The
  airlines employees do not receive a reward that
  in fact they deserved.

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• Example Slowing the Growth of Tumors
• A pharmaceutical company issued a press release
  announcing that it had filed an application with
  the Food and Drug Administration to begin
  clinical trials of an experimental drug that had
  been found to reduce the growth rate of
  pancreatic and colon cancer tumors in animal
  studies.
• Let µ denote the true mean growth rate of
  tumors for patients receiving the experimental
  drug. Set up a hypotheses testing for µ.
• Solution H0 µ mean growth rate of tumors
  for patients not taking the experimental drug
  (The drug is not effective) versus
• Ha µ lt mean growth rate of tumors
  for patients not taking the experimental drug
  (The growth rate is reduced and the drug is
  effective.)
• Type I error (Reject a true H0) Incorrectly
  conclude that the drug is effective in slowing
  the growth rate of tumors.
• Type II error (Fail to reject a false H0)
  Conclude that the experimental drug is
  ineffective when in fact it does reduce the mean
  growth rate of tumors.

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Definition Level of Significance

• The probability of a Type I error is denoted by a
  and is called the level of significance of the
  test.
• For example, a test with a .01 is said to have
  a level of significance of .01.
• The probability of a Type II error is denoted by
  ß.
• After assessing the consequences of Type I and
  Type II errors, identify the largest a that is
  tolerable for the problem.
• Dont make the a smaller than it needs to be.
• Smaller a increases ß.

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Example Blood Test for Ovarian Cancer

• A new blood test has been developed that
  appears to be able to identify ovarian cancer at
  its earliest stage. In a report issued by NCI and
  FDA the following information is given
• The test was given to 50 women known to have
  ovarian cancer, and it correctly identified all
  of them as having cancer.
• The test was given to 66 women known not to
  have ovarian cancer, and it correctly identified
  63 of them as being cancer free.
• Give an estimate of a and ß.

Solution on next slide
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Solution to the Example Blood Test for Ovarian
Cancer
H0 woman has ovarian cancer Ha woman does not
have ovarian cancer Type I error (reject H0
while it is in fact true) Believing that a
woman with ovarian cancer is cancer free. Type II
error (fail to reject H0 while it is actually
false) Believing that a woman who is actually
cancer free has ovarian cancer. Based on the
preliminary evaluation of the blood text, The
probability of a Type I error a 0/50 0. (50
women with ovarian cancer were all correctly
identified. No error.) The probability of a Type
II error ß 3/66 .046 (Among 66 women known
not to have ovarian cancer, only 3 of them were
wrongly identified as having ovarian cancer.)
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• Example Lead in Tap Water.
• Drinking water is considered unsafe if the mean
  concentration of lead is 15 ppb (parts per
  billion) or greater. The Environmental Protection
  Agency (EPA) had indicates that 6 of public
  water systems contained too much lead, and
  requires the communities to take corrective
  actions. One of the cited communities wants to
  conduct a hypotheses testing for the mean
  concentration of lead in its tap water.
• Let µ the mean concentration of lead in the
  communitys water system.
• H0 µ 15 (mean lead concentration excessive
  by EPA standard)
• versus Ha µ lt 15 (mean lead concentration at
  acceptable level)
• Type I error (Reject a true H0) To conclude that
  the water source meets EPA standards for lead
  when in fact it does not.
• Type II error (Fail to reject a false H0) To
  conclude that the water does not meet EPA
  standards when in fact it does.
• Type I error results in potentially serious
  public health risk. A small value of a such as a
  0.01 could be selected. Of course, smaller a
  increases the risk of a Type II error, which also
  has serious consequences for the community
  (losing its water source).

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10.4 Hypothesis Tests for a Population Mean

• A test statistic is the function of sample data
  on which a conclusion to reject or fail to reject
  H0 is based
• The P-value (observed significance level) is a
  measure of inconsistency between the hypothesized
  value for a population characteristic and the
  observed sample. It is the probability, assuming
  that H0 is true, of obtaining a test statistic
  value at least as inconsistent with H0 as what
  actually resulted.
• H0 should be rejected if P-value a.
• H0 should not be rejected if P-value gt a.

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Finding P-Values for a t-Test (One-Tail)
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Finding P-Values for a t-Test (Two-Tailed)

• Find P-Value for the following t test.
• Upper-tailed test, df 25, t 2.0.
• Lower-tailed test, n 28, t - 4.2.
• Two-tailed test, df 40, t 1.7.

Solutions on next slide
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Find P-Values for a t-Test Using Appendix Table
4 Tail Areas for t-Curves (page 709 page 711)

• The next slide contains a sample page of Appendix
  Table 4 (the last page). Use the table to find
  the P-values.
• Solutions to Problems 1, 2, and 3 on the
  preceding slide.
• P-Value (an upper-tailed t-test ) The area
  under the 25-df t curve to the right of 2.0
  The value in the t 2.0 row under the column
  25-df in Appendix Table 4 .028.
• P-Value (a lower-tailed t-test) The area under
  the 27-df t curve to the left of -4.2 The area
  under the 27-df t curve to the right of 4.2
  The value in the t 4.2 row under the column
  27-df in Appendix Table 4 0. (4.2 gt 4.0, the
  largest t value in the table.)
• P-Value (a two-tailed t-test) 2 The area
  under the 40-df t curve to the right of 1.7 2
  The value in the t 1.7 row under the column
  40-df in Appendix Table 4 2 .048 .096.

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t-Test for A Population Mean

• Assumption The sample size is large (generally n
  30) or at least the population distribution is
  at least approximately normal.
• Null Hypothesis H0 µ hypothesized value.
• Test statistic
• Alternative Hypothesis P-Value
• Ha µ gt hypothesized value area in upper tail
  
  
• Ha µ lt hypothesized value area in lower tail
• Ha µ ? hypothesized value sum of area in two
  tails
• Conclusion
• Reject H0 if P-value Significance level a, or
  P-value 0 (usually lt .01) even if there is no
  significance level given.

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Example Time Stands Still (or So It Seems)

• A study conducted by researchers at PSU
  investigated whether time perception, a simple
  indication of a persons ability to concentrate,
  is impaired during nicotine withdrawal. After a
  24 hours smoking abstinence, 20 smokers were
  asked to estimate how much time had passed during
  a 45-sec period. The resulting data on perceived
  elapsed time (in seconds) were shown on the
  right.
• The researchers wanted to determine
  whether smoking abstinence had a negative impact
  on time perception, causing elapsed time to be
  overestimated.
• Solution Assumptions The sample size is only
  20, we must be willing to assume that the
  population distribution of perceived times is at
  least approximately normal. The boxplot is not
  too skewed and there is no outliers, so our
  assumption is reasonable.

Continued on next slide
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Solution to the Example Time Stands Still
Let µ mean perceived elapsed time for
smokers who have abstained from smoking for 24
hours. From the data, we can find

• The researchers want to know if smoking
  abstinence caused elapsed time to be
  overestimated, so we use an upper-tailed test
  with significance level a .05.
• Null and Alternative hypotheses H0 µ 45 Ha
  µ gt 45

P-value df 20 - 1 19. Use the df 19 column
of Appendix Table 4 to find P-value area to
the right of 6.50 0 (because 6.50 gt 4.0, the
largest tabulated value, and the area to the
right of 4.0 is already 0). Conclusion Reject H0
because P-value lt a. There is convincing evidence
that the mean perceived time elapsed is
overestimated.
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Example Goofing Off at Work

• A growing concern of employers is time spent in
  activities like surfing the Internet and emailing
  friends during work hours. Suppose a CEO of a
  large corporation wants to determine if her
  employees spend less than 120 minutes a day on
  personal use of company technology. A random
  sample of 10 employees was contacted and the
  resulting data (in minutes per day) are listed
  below.

Does the data provide evidence that the
mean wasted time for the company is less than 120
min? Carry out a hypothesis test with a 0.05.
Solution Assumptions The sample size is only
10, we must be willing to assume that the
population distribution of wasted times is at
least approximately normal. Although the boxplot
reveals some skewness in the sample, there are no
outliers.
Continued on next slide
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Solution to the Example Goofing Off at Work
Let µ mean daily wasted time for
employees of the company. Summary quantities of
the data are

• The CEO wants to determine whether the mean
  wasted time for the company is less than 120
  hours, so we use a lower-tailed test with a
  .05.
• Null and Alternative hypotheses H0 µ
  120 Ha µ lt 120

P-value From the df 9 column of Appendix
Table 4, we find P-value area to the left of
-1.1 area to the right of 1.1 0.150
Conclusion Fail to reject H0 because P-value gt
a. There is no sufficient evidence to conclude
that the mean wasted time at the company is less
than 120 hours.
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Example pH Level of Water

• An chemical plant claims that the mean pH level
  of the water in a nearby river is 6.8. You
  randomly select 19 water samples and measure the
  pH value of each. The sample mean is 6.7 and
  standard deviation is 0.24. Is there enough
  evidence to reject the chemical plants claim at
  a 0.05? Assume the population is normally
  distributed.
• Solution The sample size is only 19, and
  therefore, the assumption is necessary that the
  population distribution is normal.
• To test the claim that the mean pH level is
  6.8, we use a two-tailed test with significance
  level a 0.05.

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Solution to the Example Goofing Off at Work
Let µ mean pH level in the nearby
river. The sample mean and standard are already
computed

• Null and Alternative hypotheses H0 µ 6.8 Ha
  µ ? 6.8

P-value From the df 19 - 1 18 column of
Appendix Table 4, we find P-value 2 area to
the right of 1.8 2 0.044 0.088 gt a
Conclusion Fail to reject H0 because P-value gt
a. There is no sufficient evidence at the 5
level of significance to reject the claim that
the mean pH level in the river is 6.8.
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Exercise Cricket Love

• The usual chirp rate for male field crickets
  varied around a mean of 60 chirps per second. To
  investigate if chirp rate was related to
  nutritional status, investigators fed 32 male
  crickets on the high protein diet for 8 days and
  found that the mean chirp rate for these crickets
  was 109 chirps per second with a standard
  deviation s 40. Is this convincing evidence
  that the mean chirp rate for crickets on a high
  protein diet is greater than 60 chirps per
  second? Carry out a hypothesis test with a
  significance level a .01.

Answer There is convincing evidence that the
mean chirp rate is higher for male crickets that
eat a high protein diet.