Statistics 222

1
Statistics 222

• Chapter 9
• Hypothesis Tests - Part A

2
What is a Hypothesis?

• A hypothesis is a statement which is usually
  thought to be true, and serves as a starting
  point in looking for arguments (or evidence) to
  support it.
• Here are some other definitions
• A statement of the relation between two or more
  variables.
• A supposition or assumption advanced as a basis
  for reasoning or argument, or as a guide to
  experimental investigation.
• An unproven or unverified idea or model.
• A tentative and testable guess or premise.

3
Hypotheses are derived from research questions

• Research studies start with the development of a
  research question. A testable hypothesis must be
  derived from the research question.
• Example
• Research question Does smoking cigarettes cause
  lung cancer?
• Hypothesis People who smoke have a higher
  incidence of lung cancer than people who dont
  smoke.

4
What is hypothesis Testing?

• Hypothesis testing is determining whether the
  hypothesis is true or not.
• Example To test the previously-stated
  hypothesis, you would collect a sample of
  people, some of whom smoke and others who dont.
  You would determine if the smokers have a high
  incidence of lung cancer than the non-smokers. If
  so, you conclude that the data supports the
  hypothesis.

5
Hypothesis Testing

• If the smokers have a higher incidence of lung
  cancer than the non-smokers, you can say that the
  data supports the claim that smoking causes lung
  cancer.
• When you test a hypothesis by collecting sample
  data and testing it, you cannot prove that a
  hypothesis is true or not true. You can only
  conclude that the data either supports or does
  not support the hypothesis.

6
Hypotheses in this Chapter

• In research situations, hypotheses often have to
  do with causal effects such as I believe that A
  causes B. Example I believe that smoking causes
  lung cancer.
• In this chapter, the hypotheses we develop (at
  first) have to do with a single population mean.
  For example I believe that the average
  miles-per-gallon on this vehicle is at least 24
  mpg. Or I believe that my water bottling machine
  is putting an average of 12 ounces of water in
  each bottle.

7
Hypothesis testing in this Chapter

• In this chapter, hypothesis testing is used to
  determine whether a claim about a single mean of
  a population should or should not be rejected.
• Later in the chapter, we will test hypotheses
  that are claims a single population proportions.
• In chapter 10, we test hypotheses that have to do
  with comparing two population means or
  proportions to see if there is a significant
  difference.

8
Developing a Hypothesis

• In hypothesis testing we begin by making a
  tentative assumption about a population parameter
  (e.g., a mean).
• The null hypothesis, denoted by H0, is a
  tentative assumption about a population
  parameter.
• The alternative hypothesis, denoted by Ha, is the
  opposite of what is stated in the null hypothesis.

9
Three possible forms of hypotheses

• In general, a hypothesis test about the value of
  a population mean must take one of the
  followingthree forms (where ?0 is the
  hypothesized value of the population mean).

One-tailed (lower-tail)
One-tailed (upper-tail)
Two-tailed

• The equality relational operator always appears
  in the null hypothesis.

10
Developing Null and Alternative Hypotheses

• It is not always so obvious how the Null and
  Alternative hypotheses should be formulated.
• How you derive the Ho and the Ha depends upon the
  context.
• Here are three possible contexts
• Testing a research claim
• Testing the validity of a claim
• Testing for decision-making purposes.

11
Developing a hypothesis in a research situation

• A research claim will usually involve some type
  of treatment that is being applied to
  subjects as part of an experimental study.
• The hypotheses will usually have to do with
  whether the treatment had the intended effect.
• Treatments that might be applied to human
  subjects during an experimental study might be
  behavioral modification or training programs,
  drugs, or changes in environmental conditions.

12
Example of developing a hypothesis in a research
situation

• Example A particular type of car gets 24 mpg. A
  research group developed a new fuel injection
  system designed to increase the mpg. The research
  involves installing this fuel injection system in
  several cars and then observing whether they get
  better gas mileage.
• They are looking for evidence that the fuel
  injection system increases mpg above 24.
• The car is the subject and the fuel injection
  system is the treatment, the outcome variable is
  the mpg.

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The hypotheses

• The effect that the researcher is looking for
  should be stated as the ALTERNATIVE hypothesis
• Ho ? 24 (treatment had no effect)
• Ha ? gt 24 (treatment had the intended effect)
• If the data indicate that Ho cannot be rejected,
  then they can conclude that the data supported
  the claim that the fuel injection system does NOT
  improve gas mileage.

14
Developing a hypothesis when testing the validity
of a claim

• The second type of situation where hypothesis
  formulation is appropriate is when verifying the
  validity of a claim such as a claim made by a
  manufacturer about his product.
• Other examples include advertising claim like
  this gum gives you fresher breath, or this
  cleanser will eliminate acne.

15
Example of developing a hypothesis in a research
situation

• A manufacturer of soft drinks claims that his
  bottling machines puts at least 67.6 fluid ounces
  in each bottle it fills, on average. A sample of
  2liter containers is selected and the contents
  measured in each bottle to see if the average is
  67.6 fluid ounces.

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The hypotheses

• In this situation, we assume the manufacturers
  claim is true, so the claim is the Null
  hypothesis
• Ho ? 67.6 (manufacturers claim is true)
• Ha ? lt 67.6 (manufacturers claim is not true)
• If the data taken from the sample of bottles
  indicate that Ho cannot be rejected, we will
  assume his claim is valid.

17
Developing a hypothesis in a decision-making
situation

• The third type of situation where hypothesis
  formulation is appropriate is when a
  decision-maker has to decide between two courses
  of action.
• One course of action is appropriate if Ho is
  rejected and another course of action is taken if
  Ho is not rejected.
• Use the equality sign in the null hypothesis
  rule to guide the development of your hypotheses.

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Example of developing a hypothesis in a research
situation

• A quality control manager has just received a
  shipment of nails and they are all supposed to be
  2 inches long. He takes a sample of nails from
  the boxes and measures their lengths. If the
  average length is 2, hell keep the shipment and
  if the average is not 2, hell reject the
  shipment.
• He is interested in whether the average length
  equals 2.

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The hypotheses

• Because the equality sign must be in the null
  hypothesis, we express the hypotheses in this
  form
• Ho ? 2 (shipment of nails is okay)
• Ha ? ? 2 (shipment of nails is not okay)
• If the sample results indicate that Ho cannot be
  rejected, he will accept the shipment. If the
  sample results indicate that Ho can be rejected,
  he will return the shipment.

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Question (p. 340- 1)

• The manager of a Danvers-Hilton resort hotel
  stated that the mean guest bill for a a weekend
  is 600 or less. A member of the hotels
  accounting staff noticed that the total charges
  for guest bills have been increasing in recent
  months. The accountant used a sample of weekend
  guest bills to test the managers claim.
• Which of the three situation is this?

Testing the validity of a claim
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Answer (p. 340- 1)

• What are the null and alternative hypotheses?
• What conclusion is appropriate when Ho cannot be
  rejected?
• What conclusion is appropriate when Ho can be
  rejected?

Ho ? 600 (claim is true) Ha ? gt 600
(claim is not true)
The average guest hotel bill is not above 600
The average guest hotel bill is above 600.
22
Question (p. 340- 3)

• A production line operation is designed to fill
  cartons with laundry detergent to a weight of 32
  ounces. A sample of cartons is periodically
  selected and weighed to determine whether
  under-filling or over-filling is occurring. If
  either under-filling or over-filling is
  occurring, the production line will be shut down
  and adjusted.
• Which of the three situation is this?

A decision-making situation
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Answer (p. 340- 3)

• What are the null and alternative hypotheses?
• What course of action should be taken if Ho
  cannot be rejected?
• What course of action should be taken if Ho can
  be rejected?

Ho ? 32 (production process is okay) Ha ? ?
32 (production process is not okay)
Do not shut down the production line.
Shut down the production line.
24
Types of errors that can occur during hypothesis
testing

• Type I error occurs when we reject Ho when it is
  in fact TRUE.
• Type II error occurs when we fail to reject Ho
  when it is in fact FALSE.

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Type I errors

• In a hypothesis test, a type I error occurs when
  the null hypothesis is rejected when it is in
  fact true that is, H0 is wrongly rejected.
• For example, in a clinical trial of a new drug,
  the null hypothesis might be that the new drug is
  no more effective, on average, than the current
  drug that is H0 there is no difference in
  effectiveness between the two drugs on average. A
  type I error would occur if we concluded that the
  two drugs produced different effects when in fact
  there was no difference between them.

26
Type II errors

• In a hypothesis test, a type II error occurs when
  the null hypothesis Ho, is not rejected when it
  is in fact false.
• For example, in a clinical trial of a new drug,
  the null hypothesis might be that the new drug is
  no more effective, on average, than the current
  drug that is Ho there is no difference between
  the two drugs on average. A type II error would
  occur if it was concluded that the two drugs
  produced the same effect, that is, there is no
  difference between the two drugs on average, when
  in fact they were different.

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Type I and Type II Errors
Population Condition
H0 True (m lt 12)
H0 False (m gt 12)
Conclusion
Correct Decision
Type II Error
Accept H0 (Conclude m lt 12)
Correct Decision
Type I Error
Reject H0 (Conclude m gt 12)
Type I ROT (Rejecting O when True) Type II FROF
(Fail to Reject O when False)
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An analogy

• In the US legal system, a person charged with a
  crime is innocent until proven guilty. Therefore
  the jury starts with these working hypotheses
• Ho The defendant is innocent
• Ha The defendant is guilty
• The jury examines evidence to see if Ho should be
  rejected. When the jury reaches a verdict, two
  kinds of errors could occur
• The person is innocent but found guilty
• The person is guilty but found not guilty.

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Errors that could occur when reaching a verdict

• Type I error Null is rejected when it is in fact
  true. What would be the type I error in this
  case?
• Type II error Null is NOT rejected when it is in
  fact false. What would be the type 2 error in
  this case?

If the defendant is found guilty when he is in
fact innocent.
If the defendant is found not guilty when he is
in fact guilty.
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The possible outcomesHo He is not guilty
Found not guilty
He is not guilty
Type 1 error
Found guilty
Type II error
Found not guilty
He is guilty
Found guilty
31
Type I and Type II

• The probability of a Type I error is designated
  by the Greek letter alpha (?) the probability of
  a Type II error is designated by the Greek letter
  beta (?) .
• A type I error is more serious because an
  incorrect conclusion was drawn. Actions will
  taken that shouldnt be.
• A type II error occurs when there is just not
  enough evidence to reject the null (insufficient
  evidence to convict). In research studies, type
  II errors are frequently due to sample sizes
  being too small (not enough data).

32
Avoiding Type I errors

• Because a type I error is considered to be more
  serious, and therefore more important to avoid,
  the procedure we follow as researchers is to
  control, or in other words, establish a fixed
  probability of making a type 1 error by setting
  it at some specific level such as 1, 5 or 10.
• Can we control the probability of making a type
  II error in the same way? No. The probabilities
  of making a type I and type II errors are
  inversely related the smaller the risk of one,
  the higher the risk of the other. So setting ?
  low, causes ? to be high.
• However, there are procedures for controlling the
  probability of making a type II error to a
  limited extent.

33
Question (p. 342 - 5)

• Americans spend an average of 8.6 minutes a day
  reading newspapers. A researcher believes that
  individuals in management positions spend more
  than the national average time per day reading
  newspapers. A sample of individuals in management
  positions will be selected by the researcher.
  Data on newspaper-reading times will be used to
  test the following null and alternative
  hypotheses
• Ho ? 8.6
• Ha ? gt 8.6
• What is the type 1 error in this case? What is
  the consequences of making this type of error?
• What is the type II error in this case? What is
  the consequences of making this type of error?

34
Answer (p. 342 - 5)

• Type 1 Null is rejected when it is true.
• It is wrongly concluded that individuals in
  management positions read newspapers longer than
  8.6 minutes a day when they do not.
• Type II Null is NOT rejected when it is false.
• We cannot conclude that individuals in management
  positions read newspapers longer than 8.6 minutes
  a day although they do.

35
Question (p. 342 - 7)

• Carpetland salespersons average 8000 per week in
  sales. Steve Contois, the firms VP, proposes a
  compensation plan with new selling incentives.
  Steve hopes that the results of a trial selling
  period will enable him to conclude that the
  compensation plan increases the average sales per
  salesperson.
• Develop a null and alternative hypothesis.
• What is the type 1 error in this case? What is
  the consequences of making this type of error?
• What is the type II error in this case? What is
  the consequences of making this type of error?

36
Answer (p. 342 - 7)

• Ho ? 8000
• Ha ? gt 8000
• Type 1 Null is rejected when it is true.
• It is wrongly concluded that the incentive
  program is successful when it is not.
• Type II Null is NOT rejected when it is false.
• We cannot conclude that the sales incentive
  program is successful even though it is.

37
9.3 Procedure for testing a hypothesis when ? is
known

• Recall that ? is the population standard
  deviation.
• Here are the steps
• 1. Develop the null and alternative hypothesis
• 2. Specify a significance level (?).
• 3. Gather the data.
• 4. Calculate the test statistic and perform a
  comparative test (using either the p-method or
  the critical value method).
• 5. State the conclusion to reject or not reject
  the null hypothesis and give the reason.

38
Example

• The manufacturer of Hilltop coffee claims that
  his coffee cans contain 3 lbs of coffee. One of
  the purposes for the FTC (Federal Trade
  Commission) is to verify that manufacturers
  claims are accurate.
• So an FTC guy is sent to the factory to verify
  the claim that the average weight of a can of
  Hilltop coffee is 3.0 lbs. He will take a sample
  of cans, weigh them, and see what their average
  weight is.

39
1. Develop the hypothesis

• Since this is a manufacturers claim
• Ho ? 3.0 lbs (manufacturers claim is true)
• Ha ? lt 3.0 lbs (manufacturers claim is not
  true)
• Because the Ho contains an greater than or
  less than operator, well do a one-tailed
  hypothesis test.
• If the average weight of the coffee cans in the
  sample is too far below 3 lbs, we will reject the
  null hypothesis.

40
Basis for rejecting the hypothesis

• We will assume that the hypothesis is true as an
  equality the mean of this population (?) is 3.0.
• If the sample mean is too far in the wrong
  direction, we reject Ho.
• Because Ho has a greater than operator, the
  wrong direction is left. So we do a left-tailed
  test).

.
X?
X?
X?
X?
Reject H0!
Assume 3.0
41
2. Specify the significance level ?

• Practically speaking, ? is used to provide a
  cut-off point that determines when a test
  statistic (e. g., a sample mean) is just too
  significantly different from a hypothesized
  population parameter (such as ?) to remain
  convinced that the hypothesis is true.

42
Implications of selecting ?

• Common choices for ? is are 1, 5 or 10. If the
  cost of making a type I error is really high,
  then well chose 1 if the cost of making a type
  1 error is not that high, well chose 10.
• What is the cost in this case? Recall that a type
  1 error is rejecting the null when its true. In
  this case, that would mean falsely accusing
  Hilltop of cheating the customer.
• Since false accusations can be costly, we want
  to be very certain that we dont make that
  mistake, so we will set ? at 1.

43
The role of ? in hypothesis testing

• Since we decided on 1, the implication is that
  were going over-board to make sure we dont
  reject the null too readily and falsely accuse
  Hilltop of cheating the customer.
• In other words, were taking the lowest possible
  chance of making a type 1 error.
• The only way to get ? to 0 would be to not use
  sample statistic to estimate a population
  parameter in other words, do a census not a
  survey.

44
The role of ? in hypothesis testing

• Hypothesis testing starts with the assumption
  that the null is true, in this example, that
  would be ? 3.0.
• A hypothesis test is based on the premise that if
  our sample mean is too far in the wrong direction
  (i.e. significantly different) from the
  hypothesized population mean (3.0), then the
  hypothesis is probably not true, so we reject it.
• The significance level (?) is used to calculate
  the rejection point between what is and what is
  not considered significantly lower than 3.0.

45
? Defines the rejection area
Since Ho contains , the rejection region is
only on the left or low end. Because ? is 1, the
size of the rejection region is 1 of the total
area. So if our sample mean (x) falls into this
left 1 area, we reject the hypothesis that ?
3.0.
.
We only reject the null if x falls in this end
(too small).
1
3.0
46
? is also the probability of making a type I
error.

• If our sample mean (x) turns out to be where the
  red x is, it will be cause to reject the null
  hypothesis that ?3.0. But notice that red-X is
  still inside the ?3 distribution curve.
• Therefore, we might decide to reject the null
  when x does actually belong in the ?3
  distribution curve. If that happens, we have made
  a Type 1 error.

X?
X?
X?
X?
Reject H0!
Assume 3.0
47
3. Gather the data

• The next step in hypothesis testing is to gather
  data.
• Lets say that an FTC guy shows up at the factory
  one day, pulls 36 cans of coffee off the shelf
  and weighs them to see if their average weight is
  3.0 lbs.
• It turns out that the samples average weight is
  2.92.
• We know that the population standard deviation ?
  is .18 lbs. (given).

48
Is 2.92 in the rejection area?
2.92?
2.92?
2.92?
1
3.0
49
4. Calculate the test statistic (zx)

• The test statistic is used to determine if 2.92
  is in the rejection area.
• In this case, where ? is known, z is our test
  statistic.
• Once the test statistic is calculated, we can
  plug the test statistic into either (1) the
  p-formula or (2) the critical value formula.
• The result of either formula can be used to
  determine whether to reject the null hypothesis.

50
The test statistic (zx)
x - ?0
zx
? x
2.92 3.0
-.08
-2.67
zx

.18 /?36
.03
We now know that 2.92 is 2.67 standard
deviations to the left of the mean of 3.0.
51
z is distance is expressed as the number of
standard deviations.
2.92 is 2.67 standard deviations below 3.0.
.18 / v36
-2.67
3.0
2.92
52
if 2.92s z value indicates that it is in that
left 1 area
What were trying to determine is
then this is probably the true situation ? lt
3.0
1
3.0
2.92
2.8
53
The comparative test

• There are two comparative tests that we can use
  to determine whether our sample mean (2.92) is in
  the rejection region (left 1 of area under
  curve)
• The p-value test (compares px to pc)
• The critical value test (compares zx to zc)

54
The p-value procedure

• First we calculate px. Px represents the
  probability that the null hypothesis is true
  (?3.0) given that x 2.92.
• Since probability area, The area to the left of
  the 2.92 line is equal to the probability that
  ?3.0 given that x 2.92.

?
3.0
2.92
55
P procedure step 1 Calculate px

• To find the probability that ?3.0 given that x
  2.92, we use 2.92s z-value to find a p-value.
  That p-value is the probability that ?3.0 given
  that x 2.92
• Recall that 2.92s z-value is -2.67.
• To find the area to the left of the -2.67 line
• px NORMSDIST(-2.67)
• px .0038

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P-procedure step 2 Calculate pc

• pc is the area of the rejection region.
• pc ?
• In this case, pc .01

57
P procedure step 3 Compare pc to px

• If px lt pc , reject the null hypothesis.
• Since .0038 lt .01, the null hypothesis is
  rejected.

58
We see that px (.38) lt pc (1), so the null is
rejected
1
2.92
3.0
.38
59
We conclude this ? lt 3.0
2.8?
60
5. State conclusion and reason

• We reject the null hypothesis because our
  hypothesis is too improbable to be true.
• We are rejecting the null because the probability
  that ?3.0 is less than 1. (The p-value is that
  probability when x2.92 and it was .38.)
• In other words, if it turns out that there is a
  less-than-1 probability that ? could be 3.0
  given that x2.92, its too improbable to be true.

61
The p-method (? known)
X
?
X - ?0
? x
zx
pc ?
NORMSDIST(zx) tails or 1-NORMSDIST(zx) tails
px
pc
If px lt pc, reject Ho
62
The critical value method

• Again, we are trying to determine if 2.92 is in
  the rejection region but we use a different
  method.
• Since we already have 2.92 expressed as a z-score
  (the test statistic zx of -2.67), all we need do
  now is express ? as a z-score (zc), and then
  compare the two z-scores.
• If zx lt zc ? reject null.

63
Critical Value procedure step 1 Calculate zx

• Recall that zx (the z-value of 2.92) is the test
  statistic (-2.67).

64
Critical Value procedure step 2 Calculate zc

• We transform ? into a z value
• zc NORMSINV(?)
• zc NORMSINV(.01)
• zc -2.33

65
Critical Value procedure step 3 Compare zc to zx

• If zx gt zc ? reject null
• Since 2.67 gt 2.33 ? so null is rejected when
  x 2.92.

66
We see that zx 2.67 gt zc 2.33, so the null
is rejected
1
-2.33
2.92 z-2.67
3.0
67
We conclude this ? lt 3.0
2.8?
68
5. State conclusion and reason

• We reject the null because the z-value of 2.92
  (-2.67) indicates that 2.92 is in that 1 left
  tail.
• When we use the critical value approach, the
  reason why are rejecting the null is because the
  sample mean is significantly below ?.

69
The critical value method (? known)
X
?
x - ?0
Normsinv(?)
? x
zx
zc
If zx gt zc reject Ho
70
Open the file DataSetsForCh9.xls and click on the
worksheet Hilltop Coffee to see the calculations
associated with this hypothesis test.
71
A comparison of the two methods

• It doesnt matter which one you use for a given
  mean and ?, either method will lead to the same
  conclusion.
• Now if you set ? at 5 or 10 instead of 1, that
  could change the conclusion because the cut-off
  point would be different.
• In practice, most researchers use the p-value
  approach because the p-value tells you how
  significant the results are (marginally
  significant or clearly significant).

72
The probability of Type I error

• Recall that in this example, we rejected the null
  hypothesis because the probability that it was
  true given that x2.92 was just too low to be
  true (only .38 chance it was true).
• But there is still that .38 chance that we are
  wrong. So that is why .38 is also the
  probability of a type 1 error rejecting the null
  when we should not have.

73
How would a Type I error occur?

• A type I error would occur if ? really is 3.0
  and the sample that we select turns out to be one
  of the samples in the left tail 1 of this
  distribution (area to the left of the red line).

3.0
74
But if a Type 1 error doesnt occur
Hopefully, this happened (on the right) and not
this (on the left).
Hopefully, the sample mean indicated correctly
that the population mean is below 3.0
3.0
2.8
75
Conclusion

• The hypothesis testing procedure requires us to
  make inferences from sample data about a
  population and therefore, the possibility always
  exists that we will make a wrong conclusion. Data
  supports hypotheses, it doesnt prove them.
• Type I error quantifies the probability of making
  that wrong conclusion. We attempt to minimize it
  but we cant eliminate it.

76
The Two-Tailed test

• The word tail refers to the outer wing of a
  bell curve.
• The previous example was a one-tailed test
  because the null hypothesis had an inequality
  symbol (lt) and would only be rejected if the
  sample mean weight of a coffee can was too low
  not too high. In other words, the null hypothesis
  would only be rejected if the sample mean was
  off or around the left tail of the bell curve.
• This next example is a two-tailed test where
  the null hypothesis will be rejected if the
  sample mean is too low or too high.

77
Example

• The US Golf Association establishes rules that
  manufacturers of golf equipment must meet if
  their products are to be acceptable for use in
  USGA events. A company named MaxFlight makes golf
  balls that are expected to travel an average
  distance of 295 yards.
• If their manufacturing process gets out of
  adjustment, the balls may travel too far or not
  far enough.
• The USGA comes into the factory periodically and
  tests a sample of 50 golf balls. A hypothesis
  test is conducted to see if the process has
  fallen out of adjustment.

78
1. Develop the hypothesis

• Since this is a decision-making situation, and
  action will be taken in either case, the
  hypothesis are
• Ho ? 295
• Ha ? ? 295
• If x turns out to be significantly above or below
  295, we reject the null.

79
Basis for rejecting the hypothesis
We will assume that the null hypothesis is true
the mean of this population (?) is 295. Because
Ho has an equality operator, we reject the
null hypothesis if our sample mean is too far
below or too far above 295 (a two-tailed test).
.
X?
X?
X?
X?
Reject H0!
Reject H0!
Assume 295
80
2. Specify significance level (?)

• The researcher evaluates the cost of making a
  type I error and selects .05.
• Heres the rule for a two-tailed test any sample
  mean that falls in the right ?/2 tail area OR in
  the left ?/2 of the distribution is considered
  significantly above or below 295.

81
? defines the rejection areas(For a two-tailed
test, ? is split between the two-tails)
2.5
2.5
295
82
3. Gather the data

• The sample size is 50
• Therefore (n gt 30) we can assume that the
  distribution of sample means is normal.
• The population ? is given 12
• Therefore the standard error of the means (?x) is
  12 / v50 or 1.7
• The sample mean turns out to be 297.6.

83
Is 297.6 in either one of the rejection areas?
297.6?
297.6?
297.6?
2.5
2.5
295
84
4. Calculate the test statistic (z)
x - ?o
297.6 295
2.6
zx

1.53

? x
12 /?50
1.7
We now know that 297.6 is 1.53 standard
deviations to the right of the mean.
85
Using the p-method for a two-tail test

• Now you would think that the next thing we want
  to know What is the probability of getting a
  297.6 if the ?295?
• But when its a two-tailed test, we actually ask,
  what is the probability of getting a value 1.53
  standard deviations above 295 or 1.53 standard
  deviations below 295?
• So we need to find the probability of getting a
  value 1.53 standard deviations above 295 and then
  multiply it by the number of tails to get our
  p-value.

86
The p-value method for a 2-tail testWhat is px
when zx 1.53?
Probability Area Since our initial z-value is
positive, first we find the area to the right of
the 1.53 line and then we double it to find the
area that is in both the right area and to that
left of the -1.53 line.
?
?
-1.53
297.6 z1.53
295
87
P procedure step 1 Calculate px

• We use the test statistic (zx) which is 1.53 to
  find the area to the right of the 297.6 line then
  multiply it by tails so that it will be double
  if tails is 2
• px (1 NORMSDIST(1.53)) tails
• px (1 - .9370) 2
• px .0630 2
• px .1260

For a two-tailed test, Px is doubled.
88
P-procedure step 2 Calculate pc

• Pc is the area of the rejection region.
• pc ?
• In this case, pc .05

89
P procedure step 3 Compare pc to px

• If px lt pc , reject the null hypothesis.
• Since .1260 gt .05, the null hypothesis cannot be
  rejected.

90
We see that .1260 gt .05, so the null cannot be
rejected.
6.30
6.30
2.5
2.5
297.6 z1.53
-1.53
295
91
5. Conclusion and reason

• Do not reject the null hypothesis because, based
  on the sample mean of 297.6, the probability that
  ?295 is 12.6, which is greater than 5.

92
The critical value methodfor a two-tail test

• Again, we are trying to determine if 297.6 is in
  the rejection region but we use a different
  method.
• Since we already have 297.6 expressed as a
  z-score (the test statistic zx of 1.53 so all we
  need do now is express ? as a z-score (zc), and
  then compare the two z-scores.
• For a two-tailed test, zc is halved because the
  rejection region is divided between the two
  tails.
• If zx lt zc ? reject null.

93
Critical value procedure step 1 Calculate zx

• Recall that zx (the z-value of 297.6) is the test
  statistic (1.53).

94
Critical Value procedure step 2 Calculate zc

• Recall that we transform ? into a z value.
• But because this is a two-tailed test, we use ? /
  2 (because one single rejection region is 2.5 -
  not 5)
• zc NORMSINV(? / tails)
• zc NORMSINV(.05 / 2)
• zc -1.96

For a two-tailed test, zc is halved.
95
Critical value procedure step 3 Compare zc to zx

• If zx gt zc ? reject null
• Since 1.53 lt 1.96 ? null is NOT rejected when
  x 297.6.

96
We see that 1.53 lt 1.96, so the null cannot
be rejected.
2.5
2.5
1.96
-1.96
297.6 z1.53
295
97
We cannot reject this ?295
295
98
5. Conclusion and reason

• Do not reject the null hypothesis that ? 295
  feet because the sample mean is not significantly
  different than this hypothesized ?.

99
Open the file DataSetsForCh9.xls and click on the
worksheet Golf Balls to see the calculations
associated with this hypothesis test.
This additional row/calculation was added
This formula was updated.
100
Practice (p. 357 21)

• Fowle Marketing Research, Inc. bases charges to a
  client on the assumption that telephone surveys
  can be completed in a mean time of 15 minutes or
  less. If a longer mean survey time is necessary,
  a premium rate is charged. Suppose a sample of 35
  surveys for a particular client shows a sample
  mean of 17 minutes. Use ? 4 minutes. Should
  this client be charged more?

101
Hypothesis

• This situation is most like a testing the
  validity of a claim because it only calls for
  action if Ho is rejected. Therefore
• Ho ? 15 (the company is claiming)
• Ha ? gt 15
• If 17 is significantly above 15, then well have
  the evidence to reject the null and charge this
  client more money.

102
The rejection region
Since Ho ? 15, the rejection region is on the
right (any x-value that is too high)
15
103
Open the file DataSetsForCH9.xls and click on the
worksheet Fowle Mktg
104
Enter the relation sign, hypothesized mean,
significance, and number of tails B5 ?
(relation sign) C5 15 (?0) C6 .01
(significance) C7 1 ( of tails)
105
Gather Data

• A sample of 35 was taken
• The mean for the sample (x) was 17
• We are told that the population standard
  deviation ? is 4 minutes.

106
Enter the population SD, sample size, and sample
mean C8 4 (?) C9 35 (n) C10 17 (x)
107
Is 17 in the rejection area?
17?
17?
17?
1
15
17
108
Calculate the test statistic (zx)
x - ? x
17 15
zx
2.958

? x
4/?35
109
First, calculate the sample distributions
standard deviation C11 C8 / sqrt(C9)
110
Now, calculate the z-value for 17 (zx) C12
(C10-C5) / C11
17 is almost 3 standard deviations above 15.
111
P procedure step 1 Calculate px

• px is the probability of getting 17 or higher
  when ?15.
• We use the test statistic (zx) which is 2.958 to
  find the area to the right of the 17 line
• px (1 NORMSDIST(2.958)) tails
• px (1 - .9985) 1
• px .0015 1
• px .0015

112
Calculate the areas under the curve on both the
left and right of the sample mean value (17) on
the distribution whose ? 15 C15
normsdist(C14) C16 1 C15
Area to left of 17
Area to right of 17
113
Determine which area is relevant to this
hypothesis test and select that area. (If z gt 0,
we want the right side, if z lt 0, we want the
left side) C17 (C12 lt 0, C15, C16)
114
Multiply the probability times the number of
tails C18 C17 C7
115
P-procedure step 2 Calculate pc

• pc is the area of the rejection region.
• pc ?
• In this case, pc .01

116
Enter the value of ? for Pc C19 C6
117
P procedure step 3 Compare pc to px

• If px lt pc , reject the null hypothesis.
• Since .0015 lt .01, the null hypothesis should be
  rejected.

118
If Px lt Pc reject the null hypothesis C20
if(C18ltC19, reject null, do not reject null)
119
P method Reject the null?
Using the p-method, we find that .0015 lt .01 so
reject the null hypothesis because its too
improbable to be true.
1
.15
17
15
120
Conclusion

• Given that the sample average time was 17
  minutes, the hypothesis that the average time to
  complete a survey is 15 minutes or less is too
  improbable to be true.
• Fowle Marketing Research should charge this
  customer more money because the average time it
  takes to do their surveys (17 minutes) is
  significantly above the overall average time (15
  minutes) on which their current fees are based.

121
Critical value procedure step 1 Calculate zx

• Recall that zx (the z-value of 17) is the test
  statistic (2.958).

122
Re-state zx C23 C12
123
Critical Value procedure step 2 Calculate zc

• Recall that we transform ? into a z value.
• We use ? as the p-value but for only one
  rejection region. To obtain zc, we divide ? by
  the number of tails
• zc NORMSINV(? / tails)
• zc NORMSINV(.01 / 1)
• zc -2.326

124
Calculate zc using normsinv( ). For the p-value,
divide ? by the number of tails to get the area
of one rejection region C24 normsinv(C6/C7)
125
Critical value procedure step 3 Compare zc to zx

• If zx gt zc ? reject null
• Since 2.326 lt 2.96 ? null is rejected when x
  17.

126
If zx gt zc, reject the null hypothesis C25
if(abs(C23) gt abs(C24), Reject null, Do not
reject null)
127
Conclusion

• The null hypothesis should be rejected because
  the sample average time of 17 minutes is
  significantly above the hypothesized mean time of
  15 minutes.

128
Homework 3

• Hypotheses formulation
• 2 (p. 340) and 3 (p. 340)
• Identify Type I and II errors
• 6 (p. 342), 8 (p. 343),
• Hypotheses testing
• 17 (p. 357), 20 (p. 357)