Chapter 8: Introduction to Statistical Inferences

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Chapter 8 Introduction to Statistical Inferences
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Chapter Goals

• Learn the basic concepts of estimation and
  hypothesis testing.
• Consider questions about a population mean using
  two methods that assume the population standard
  deviation is known.
• Consider what value or interval of values can we
  use to estimate a population mean?
• Consider is there evidence to suggest the
  hypothesized mean is incorrect?

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8.1 The Nature of Estimation

• Discuss estimation more precisely.
• What makes a statistic good?
• Assume the population standard deviation, s, is
  known throughout this chapter.
• Concentrate on learning the procedures for making
  statistical inferences about a population mean m.

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• Point Estimate for a Parameter
• The value of the corresponding statistic.
• Example
• is a point estimate (single number
  value) for the mean m of the sampled population.
• Problem
• How good is the point estimate? Is it high? Or
  low? Would another sample yield the same
  result?
• Note
• The quality of an estimation procedure is
  enhanced if the sample statistic is both less
  variable and unbiased.

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• Unbiased Statistic
• A sample statistic whose sampling distribution
  has a mean value equal to the value of the
  population parameter being estimated. A
  statistic that is not unbiased is a biased
  statistic.
• Example
• The figures on the next slide illustrate the
  concept of being unbiased and the effect of
  variability on a point estimate.
• Assume A is the parameter being estimated.

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Negative bias Under estimate High variability
Unbiased On target estimate
Positive bias Over estimate Low variability
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• Note
• 1. The sample mean, , is an unbiased statistic
  because the mean value of the sampling
  distribution is equal to the population mean
• 2. Sample means vary from sample to sample. We
  dont expect the sample mean to exactly equal the
  population mean m.
• 3. We do expect the sample mean to be close to
  the population mean.
• 4. Since closeness is measured in standard
  deviations, we expect the sample mean to be
  within 2 standard deviations of the population
  mean.

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• Interval Estimate
• An interval bounded by two values and used to
  estimate the value of a population parameter.
  The values that bound this interval are
  statistics calculated from the sample that is
  being used as the basis for the estimation.
• Level of Confidence 1 - a
• The probability that the sample to be selected
  yields an interval that includes the parameter
  being estimated.
• Confidence Interval
• An interval estimate with a specified level of
  confidence.

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• Summary
• To construct a confidence interval for a
  population mean m, use the CLT.
• Use the point estimate as the central value
  of an interval.
• Since the sample mean ought to be within 2
  standard deviations of the population mean (95
  of the time), we can find the bounds to an
  interval centered at
• The level of confidence for the resulting
  interval is approximately 95, or 0.95.
• We can be more accurate in determining the level
  of confidence.

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• Illustration
• The interval
• is an approximate 95 confidence interval for the
  population mean m.

Distribution of
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8.2 Estimation of Mean m(s Known)

• Formalize the interval estimation process as it
  applies to estimating the population mean m based
  on a random sample.
• Assume the population standard deviation s is
  known.
• The assumptions are the conditions that need to
  exist in order to correctly apply a statistical
  procedure.

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• The assumption for estimating the mean m using a
  known s
• The sampling distribution of has a normal
  distribution.
• Assumption satisfied by1. Knowing that the
  sampling population is normally distributed, or
• 2. Using a large enough random sample (CLT).
• Note The CLT may be applied to smaller samples
  (for example n 15) when there is evidence to
  suggest a unimodal distribution that is
  approximately symmetric. If there is evidence
  of skewness, the sample size needs to be much
  larger.

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• A (large sample) 1 - a confidence interval for m
  is found by
• Note
• 1. is the point estimate and the center point
  of the confidence interval.
• 2. z(a/2) confidence coefficient, the number of
  multiples of the standard error needed to
  construct an interval estimate of the correct
  width to have a level of confidence 1 - a.

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• 3. standard error of the mean.
• The standard deviation of the distribution of
• 4. maximum error of
  estimate E.
• One-half the width of the confidence interval
  (the product of the confidence coefficient and
  the standard error).
• 5. lower
  confidence limit (LCL).
• upper
  confidence limit (UCL).

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• The Confidence Interval A Five-Step Model
• 1. Describe the population parameter of concern.
• 2. Specify the confidence interval criteria.
• a. Check the assumptions.
• b. Identify the probability distribution and the
  formula to
• be used.
• c. Determine the level of confidence, 1 - a.
• 3. Collect and present sample information.
• 4. Determine the confidence interval.
• a. Determine the confidence coefficient.
• b. Find the maximum error of estimate.
• c. Find the lower and upper confidence limits.
• 5. State the confidence interval.

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• Example The weights of full boxes of a certain
  kind of cereal are normally distributed with a
  standard deviation of 0.27 oz. A sample of 18
  randomly selected boxes produced a mean weight of
  9.87 oz. Find a 95 confidence interval for the
  true mean weight of a box of this cereal.
• Solution
• 1. Describe the population parameter of concern.
• The mean, m, weight of all boxes of this cereal.
• 2. Specify the confidence interval criteria.
• a. Check the assumptions.
• The weights are normally distributed. The
  distribution of is normal.
• b. Identify the probability distribution and
  formula to be used.
• Use the standard normal variable z with s
  0.27.

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• c. Determine the level of confidence, 1 - a.
• The question asks for 95 confidence 1 - a
  0.95.
• 3. Collect and present information.
• The sample information is given in the statement
  of the
• problem.
• Given
• 4. Determine the confidence interval.
• a. Determine the confidence coefficient.
• The confidence coefficient is found using Table
  4B

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• b. Find the maximum error of estimate.
• Use the maximum error part of the formula for a
  CI.
• c. Find the lower and upper confidence limits.
• Use the sample mean and the maximum error
• 5. State the confidence interval.
• 9.75 to 10.00 is a 95 confidence interval for
  the true mean weight, m, of cereal boxes.

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• Example A random sample of the test scores of
  100 applicants for clerk-typist positions at a
  large insurance company showed a mean score of
  72.6. Determine a 99 confidence interval for
  the mean score of all applicants at the insurance
  company. Assume the standard deviation of test
  scores is 10.5.
• Solution
• 1. Parameter of concern the mean test score, m,
  of all applicants at the insurance company.
• 2. Confidence interval criteria.
• a. Assumptions The distribution of the
  variable, test score, is not known. However,
  the sample size is large enough (n 100) so
  that the CLT applies.
• b. Probability distribution standard normal
  variable z with s 10.5.

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• c. The level of confidence 99, or 1 - a
  0.99.
• 3. Sample information.
• Given n 100 and 72.6.
• 4. The confidence interval.
• a. Confidence coefficient
• b. Maximum error
• c. The lower and upper limits
• 5. Confidence interval With 99 confidence we
  can say,
• The mean test score is between 69.9 and 75.3.
  
• 69.9 to 75.3 is a 99 confidence interval for
  the true
• mean test score.

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• Note The confidence is in the process.
• 95 confidence means if we conduct the
  experiment over and over, and construct lots of
  confidence intervals, then 95 of the confidence
  intervals will contain the true mean value m.
• Sample Size
• Problem Find the sample size necessary in order
  to obtain a specified maximum error and level of
  confidence (assume the standard deviation is
  known).
• Solve this expression for n

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• Example Find the sample size necessary to
  estimate a population mean to within .5 with 95
  confidence if the standard deviation is 6.2.
• Solution
• Therefore, n 591.
• Note When solving for sample size n, always
  round up to the next largest integer. (Why?)

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8.3 The Nature of Hypothesis Testing

• Formal process for making an inference.
• Consider many of the concepts of a hypothesis
  test and look at several decision-making
  situations.
• The entire process starts by identifying
  something of concern and then formulating two
  hypotheses about it.

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• Hypothesis
• A statement that something is true.
• Statistical Hypothesis Test
• A process by which a decision is made between two
  opposing hypotheses. The two opposing hypotheses
  are formulated so that each hypothesis is the
  negation of the other. (That way one of them is
  always true, and the other on is always false.)
  Then one hypothesis is tested in hopes that it
  can be shown to be a very improbable occurrence
  thereby implying the other hypothesis is the
  likely truth.

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• There are two hypotheses involved in making a
  decision.
• Null Hypothesis, H0
• The hypothesis to be tested. Assumed to be true.
  Usually a statement that a population parameter
  has a specific value. The starting point for
  the investigation.
• Alternative Hypothesis, Ha
• A statement about the same population parameter
  that is used in the null hypothesis. Generally
  this is a statement that specifies the population
  parameter has a value different, in some way,
  from the value given in the null hypothesis. The
  rejection of the null hypothesis will imply the
  likely truth of this alternative hypothesis.

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• Note
• 1. Basic idea proof by contradiction.
• Assume the null hypothesis is true and look for
  evidence to suggest that it is false.
• 2. Null hypothesis the status quo.
• A statement about a population parameter that is
  assumed to be true.
• 3. Alternative hypothesis also called the
  research hypothesis.
• Generally, what you are trying to prove.
• We hope experimental evidence will suggest the
  alternative hypothesis is true by showing the
  unlikeliness of the truth of the null hypothesis.
  

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• Example Suppose you are investigating the
  effects of a new pain reliever. You hope the new
  drug relieves minor muscle aches and pains longer
  than the leading pain reliever. State the null
  and alternative hypotheses.
• Solution
• H0 The new pain reliever is no better than the
  leading pain reliever.
• Ha The new pain reliever lasts longer than the
  leading pain reliever.

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• Example You are investigating the presence of
  radon in homes being built in a new development.
  If the mean level of radon is greater than 4 then
  send a warning to all home owners in the
  development. State the null and alternative
  hypotheses.
• Solution
• H0 The mean level of radon for homes in the
  development is 4 (or less).
• Ha The mean level of radon for homes in the
  development is greater than 4.

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• Hypothesis test outcomes
• Type A correct decision
• Null hypothesis true, decide in its favor.
• Type B correct decision
• Null hypothesis false, decide in favor of
  alternative hypothesis.
• Type I error
• Null hypothesis true, decide in favor of
  alternative hypothesis.
• Type II error
• Null hypothesis false, decide in favor of null
  hypothesis.

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• Example A calculator company has just received a
  large shipment of parts used to make the screens
  on graphing calculators. They consider the
  shipment acceptable if the proportion of
  defective parts is 0.01 (or less). If the
  proportion of defective parts is greater than
  0.01 the shipment is unacceptable and returned to
  the manufacturer. State the null and alternative
  hypotheses, and describe the four possible
  outcomes and the resulting actions that would
  occur for this test.
• Solution
• H0 The proportion of defective parts is 0.01 (or
  less).
• Ha The proportion of defective parts is greater
  than 0.01.

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• Fail to Reject H0
• Null Hypothesis Is True
• Type A correct decision.
• Truth of situation The proportion of defective
  parts is 0.01 (or less).
• Conclusion It was determined that the proportion
  of defective parts is 0.01 (or less).
• Action The calculator company received parts
  with an acceptable proportion of defectives.

Null Hypothesis Is False Type II error. Truth
of situation The proportion of defective parts
is greater than 0.01. Conclusion It was
determined that the proportion of defective parts
is 0.01 (or less). Action The calculator company
received parts with an unacceptable proportion of
defectives.
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Null hypothesis is false Type B correct
decision. Truth of situation The proportion of
defectives is greater than 0.01. Conclusion It
was determined that the proportion of defectives
is greater than 0.01. Action Send the shipment
back to the manufacturer. The proportion of
defectives is unacceptable.

• Reject H0
• Null hypothesis is true
• Type I error.
• Truth of situation The proportion of defectives
  is 0.01 (or less).
• Conclusion It was determined that the proportion
  of defectives is greater than 0.01.
• Action Send the shipment back to the
  manufacturer. The proportion of defectives is
  unacceptable.

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• Note
• 1. The type II error sometimes results in what
  represents a lost opportunity.
• 2. Since we make a decision based on a sample,
  there is always the chance of making an error.
• Probability of a type I error a.
• Probability of a type II error b.

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• Note
• 1. Would like a and b to be as small as possible.
• 2. a and b are inversely related.
• 3. Usually set a (and dont worry too much about
  b. Why?)
• 4. Most common values for a and b are 0.01 and
  0.05.
• 5. 1 - b the power of the statistical test.
• A measure of the ability of a hypothesis test to
  reject a false null hypothesis.
• 6. Regardless of the outcome of a hypothesis
  test, we never really know for sure if we have
  made the correct decision.

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• Interrelationship between the probability of a
  type I error (a), the probability of a type II
  error (b), and the sample size (n).

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• Level of Significance a
• The probability of committing the type I error.
• Test Statistic
• A random variable whose value is calculated from
  the sample data and is used in making the
  decision fail to reject H0 or reject H0.
• Note
• 1. The value of the test statistic is used in
  conjunction with a decision rule to determine
  fail to reject H0 or reject H0.
• 2. The decision rule is established prior to
  collecting the data and specifies how you will
  reach the decision.

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• The Conclusion
• a. If the decision is reject H0, then the
  conclusion should be worded something like,
  There is sufficient evidence at the a level of
  significance to show that . . . (the meaning of
  the alternative hypothesis).
• b. If the decision is fail to reject H0, then the
  conclusion should be worded something like,
  There is not sufficient evidence at the a level
  of significance to show that . . . (the meaning
  of the alternative hypothesis).
• Note
• 1. The decision is about H0.
• 2. The conclusion is a statement about Ha.
• 3. There is always the chance of making an error.

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8.4 Hypothesis Test of Mean m (s known) A
Probability-Value Approach

• The concepts and much of the reasoning behind
  hypothesis tests are given in the previous
  sections.
• Formalize the hypothesis test procedure as it
  applies to statements concerning the mean m of a
  population (s known) a probability-value
  approach.

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• The assumption for hypothesis tests about a mean
  m using a known s
• The sampling distribution of has a normal
  distribution.
• Recall
• 1. The distribution of has mean m.
• 2. The distribution of has standard deviation
• Hypothesis test
• 1. A well-organized, step-by-step procedure used
  to make a decision.
• 2. Probability-value approach (p-value approach)
  a
• procedure that has gained popularity in recent
  years.
• Organized into five steps.

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• The Probability-Value Hypothesis Test A
  Five-Step Procedure
• 1. The Set-Up
• a. Describe the population parameter of concern.
• b. State the null hypothesis (H0) and the
  alternative hypothesis (Ha).
• 2. The Hypothesis Test Criteria
• a. Check the assumptions.
• b. Identify the probability distribution and the
  test statistic formula to be used.
• c. Determine the level of significance, a.
• 3. The Sample Evidence
• a. Collect the sample information.
• b. Calculate the value of the test statistic.
• 4. The Probability Distribution
• a. Calculate the p-value for the test statistic.
• b. Determine whether or not the p-value is
  smaller than a.
• 5. The Results
• a. State the decision about H0.
• b. State a conclusion about Ha.

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• Example A company advertises the net weight of
  its cereal is 24 ounces. A consumer group would
  like to check this claim. They cannot check
  every box of cereal, so a sample of cereal boxes
  will be examined. A decision will be made about
  the true mean weight based on the sample mean.
  State the consumer groups null and alternative
  hypotheses. Assume s .2.
• Solution
• 1. The Set-Up
• a. Describe the population parameter of concern.
• The population parameter of interest is the
  mean m, the mean weight of the cereal boxes.

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• b. State the null hypothesis (H0) and the
  alternative hypothesis (Ha).
• Formulate two opposing statements concerning m.
• H0 m 24 ( ) (the mean is at least 24)
• Ha m lt 24 (the mean is less than 24)
• Note
• The trichotomy law from algebra states that two
  numerical values must be related in exactly one
  of three possible relationships lt, , or gt. All
  three of these possibilities must be accounted
  for between the two opposing hypotheses in order
  for the hypotheses to be negations of each other.

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• Possible Statements of Null and Alternative
  Hypotheses
• Note
• 1. The null hypothesis will be written with just
  the equal sign (a value is assigned).
• 2. When equal is paired with less than or greater
  than, the combined symbol is written beside the
  null hypothesis as a reminder that all three
  signs have been accounted for in these two
  opposing statements.

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• Example An automobile manufacturer claims a new
  model gets at least 27 miles per gallon. A
  consumer groups disputes this claim and would
  like to show the mean miles per gallon is lower.
  State the null and alternative hypotheses.
• Solution H0 m 27 (³) and Ha m lt 27
• Example A freezer is set to cool food to .
  If the temperature is higher, the food could
  spoil, and if the temperature is lower, the
  freezer is wasting energy. Random freezers are
  selected and tested as they come off the assembly
  line. The assembly line is stopped if there is
  any evidence to suggest improper cooling. State
  the null and alternative hypotheses.
• Solution H0 m 10 and Ha m ¹ 10

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• Common Phrases and Their Negations

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• Example (continued) Weight of cereal boxes.
• Recall H0 m 24 (³) (at least 24) Ha m lt
  24 (less than 24)
• 2. The Hypothesis Test Criteria
• a. Check the assumptions.
• The weight of cereal boxes is probably mound
  shaped. A sample size of 40 should be
  sufficient for the CLT to apply. The sampling
  distribution of the sample mean can be expected
  to be normal.
• b. Identify the probability distribution and the
  test statistic to be used.
• To test the null hypothesis, ask how many
  standard deviations away from m is the sample
  mean.

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• c. Determine the level of significance.
• Consider the four possible outcomes and their
  consequences. Let a 0.05.
• 3. The Sample Evidence
• a. Collect the sample information.
• A random sample of 40 cereal boxes is examined.
• b. Calculate the value of the test statistic.
  (s .2)
• 4. The Probability Distribution
• a. Calculate the p-value for the test statistic.

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• Probability-Value, or p-Value
• The probability that the test statistic could be
  the value it is or a more extreme value (in the
  direction of the alternative hypothesis) when the
  null hypothesis is true (Note the symbol P will
  be used to represent the p-value, especially in
  algebraic situations.)

P
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• b. Determine whether or not the p-value is
  smaller than a.
• The p-value (0.0571) is greater than a (0.05).
• 5. The Results
• Decision Rule
• a. If the p-value is less than or equal to the
  level of significance a, then the decision must
  be to reject H0.
• b. If the p-value is greater than the level of
  significance a, then the decision must be to fail
  to reject H0.
• a. State the decision about H0.
• Decision about H0 Fail to reject H0.
• b. Write a conclusion about Ha.
• There is not sufficient evidence at the 0.05
  level of significance to show that the mean
  weight of cereal boxes is less than 24 ounces.

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• Note
• 1. If we fail to reject H0, there is no evidence
  to suggest the null hypothesis is false. This
  does not mean H0 is true.
• 2. The p-value is the area, under the curve of
  the probability distribution for the test
  statistic, that is more extreme than the
  calculated value of the test statistic.
• 3. There are 3 separate cases for p-values. The
  direction (or sign) of the alternative hypothesis
  is the key.

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• Finding p-values
• 1. Ha contains gt (Right tail)
• p-value P(z gt z)
• 2. Ha contains lt (Left tail)
• p-value P(z lt z)
• 3. Ha contains (Two-tailed)
• p-value
• P(z lt -z) P(z gt z)

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• Example The mean age of all shoppers at a local
  jewelry store is 37 years (with a standard
  deviation of 7 years). In an attempt to attract
  older adults with more disposable income, the
  store launched a new advertising campaign.
  Following the advertising, a random sample of 47
  shoppers showed a mean age of 39.3. Is there
  sufficient evidence to suggest the advertising
  campaign has succeeded in attracting older
  customers?
• Solution
• 1. The Set-Up
• a. Parameter of concern the mean age, m, of all
  shoppers.
• b. The hypotheses
• H0 m 37 ()
• Ha m gt 37

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• 2. The Hypothesis Test Criteria
• a. The assumptions The distribution of the age
  of shoppers is unknown. However, the sample
  size is large enough for the CLT to apply.
• b. The test statistic The test statistic will
  be z.
• c. The level of significance none given. We
  will find a p- value.
• 3. The Sample Evidence
• a. Sample information
• b. Calculated test statistic

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• 4. The Probability Distribution
• a. The p-value
• b. Determine whether or not the p-value is
  smaller than a.
• A comparison is not possible, no a given.
• 5. The Results
• Because the p-value is so small (P lt 0.05),
  there is evidence to suggest the mean age of
  shoppers at the jewelry store is greater than 37.

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• The idea of the p-value is to express the degree
  of belief in the null hypothesis
• 1. When the p-value is minuscule (like 0.0001),
  the null
• hypothesis would be rejected by everyone because
  the
• sample results are very unlikely for a true H0.
• 2. When the p-value is fairly small (like 0.01),
  the evidence
• against H0 is quite strong and H0 will be
  rejected by many.
• 3. When the p-value begins to get larger (say,
  0.02 to 0.08),
• there is too much probability that data like the
  sample
• involved could have occurred even if H0 were
  true, and the
• rejection of H0 is not an easy decision.
• 4. When the p-value gets large (like 0.15 or
  more), the data is
• not at all unlikely if the H0 is true, and no
  one will reject
• H0.

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• Advantages of p-value approach
• 1. The results of the test procedure are
  expressed in terms of a continuous probability
  scale from 0.0 to 1.0, rather than simply on a
  reject or fail to reject basis.
• 2. A p-value can be reported and the user of the
  information can decide on the strength of the
  evidence as it applies to his/her own situation.
• 3. Computers can do all the calculations and
  report the p-value, thus eliminating the need for
  tables.
• Disadvantage
• Tendency for people to put off determining the
  level of significance.

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• Example The active ingredient for a drug is
  manufactured using fermentation. The standard
  process yields a mean of 26.5 grams (assume s
  3.2). A new mixing technique during fermentation
  is implemented. A random sample of 32 batches
  showed a sample mean 27.1. Is there any evidence
  to suggest the new mixing technique has changed
  the yield?
• Solution
• 1. The Set-Up
• a. The parameter of interest is the mean yield
  of active ingredient, m.
• b. The null and alternative hypotheses
• H0 m 26.5
• Ha m ¹ 26.5

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• 2 The Hypothesis Test Criteria
• . a. Assumptions A sample of size 32 is large
  enough to satisfy the CLT.
• b. The test statistic z
• c. The level of significance find a p-value.
• 3. The Sample Evidence
• a. From the sample
• b. The calculated test statistic

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• 4. The Probability Distribution
• a. The p-value
• b. The p-value is large. There is no a given in
  the statement of the problem.
• Note Suppose we took repeated samples of size
  32.
• 1. What results would you expect?
• 2. What does the p-value really measure?

60

• 5. The Results
• Because the p-value is large (P 0.2892), there
  is no evidence to suggest the new mixing
  technique has changed the mean yield.

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8.5 Hypothesis Test of mean m (s known) A
Classical Approach

• Concepts and reasoning behind hypothesis testing
  given in previous section.
• Formalize the hypothesis test procedure as it
  applies to statements concerning m of a
  population with known s a classical approach.

62

• The assumption for hypothesis tests about mean m
  using a known s
• The sampling distribution of has a normal
  distribution.
• Recall
• 1. The distribution of has mean m.
• 2. The distribution of has standard deviation
  
• Hypothesis test
• 1. A well-organized, step-by-step procedure used
  to make a decision.
• 2. The classical approach is the hypothesis test
  process that has enjoyed popularity for many
  years.

63

• The Classical Hypothesis Test A Five-Step
  Procedure
• 1. The Set-Up
• a. Describe the population parameter of concern.
• b. State the null hypothesis (H0) and the
  alternative hypothesis (Ha).
• 2. The Hypothesis Test Criteria
• a. Check the assumptions.
• b. Identify the probability distribution and the
  test statistic to be used.
• c. Determine the level of significance, a.
• 3. The Sample Evidence
• a. Collect the sample information.
• b. Calculate the value of the test statistic.
• 4. The Probability Distribution
• a. Determine the critical region(s) and critical
  value(s).
• b. Determine whether or not the calculated test
  statistic is in the critical region.
• 5. The Results
• a. State the decision about H0.
• b. State the conclusion about Ha.

64

• Example A company advertises the net weight of
  its cereal is 24 ounces. A consumer group would
  like to check this claim. They cannot check
  every box of cereal, so a sample of cereal boxes
  will be examined. A decision will be made about
  the true mean weight based on the sample mean.
  State the consumer groups null and alternative
  hypotheses. Assume s .2.
• Solution
• 1. The Set-Up
• a. Describe the population parameter of concern.
• The population parameter of interest is the
  mean, m, the mean weight of the cereal boxes.

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• b. State the null hypothesis (H0) and the
  alternative hypothesis (Ha).
• Formulate two opposing statements concerning
  the m.
• H0 m 24 ( ) (the mean is at least 24)
• Ha m lt 24 (the mean is less than 24)
• Note
• The trichotomy law from algebra states that two
  numerical values must be related in exactly one
  of three possible relationships lt, , or gt. All
  three of these possibilities must be accounted
  for between the two opposing hypotheses in order
  for the hypotheses to be negations of each other.

66

• Possible Statements of Null and Alternative
  Hypotheses
• Note
• 1. The null hypothesis will be written with just
  the equal sign (a value is assigned).
• 2. When equal is paired with less than or greater
  than, the combined symbol is written beside the
  null hypothesis as a reminder that all three
  signs have been accounted for in these two
  opposing statements.

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• Example An automobile manufacturer claims a new
  model gets at least 27 miles per gallon. A
  consumer groups disputes this claim and would
  like to show the mean miles per gallon is lower.
  State the null and alternative hypotheses.
• Solution H0 m 27 (³) and Ha m lt 27
• Example A freezer is set to cool food to .
  If the temperature is higher, the food could
  spoil, and if the temperature is lower, the
  freezer is wasting energy. Random freezers are
  selected and tested as they come off the assembly
  line. The assembly line is stopped if there is
  any evidence to suggest improper cooling. State
  the null and alternative hypotheses.
• Solution H0 m 10 and Ha m ¹ 10

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• Common Phrases and Their Negations

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• Example (continued) Weight of cereal boxes.
• Recall H0 m 24 (³) (at least 24) Ha m lt
  24 (less than 24)
• 2. The Hypothesis Test Criteria
• a. Check the assumptions.
• The weight of cereal boxes is probably mound
  shaped. A sample size of 40 should be
  sufficient for the CLT to apply. The sampling
  distribution of the sample mean can be expected
  to be normal.
• b. Identify the probability distribution and the
  test statistic to be used.
• To test the null hypothesis, ask how many
  standard deviations away from m is the sample
  mean.

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• c. Determine the level of significance.
• Consider the four possible outcomes and their
  consequences. Let a 0.05.
• 3. The Sample Evidence
• a. Collect the sample information.
• A random sample of 40 cereal boxes is examined.
• b. Calculate the value of the test statistic.
  (s .2)

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• 4. The Probability Distribution
• a. Determine the critical region(s) and critical
  value(s).
• Critical Region
• The set of values for the test statistic that
  will cause us to reject the null hypothesis. The
  set of values that are not in the critical region
  is called the noncritical region (sometimes
  called the acceptance region).
• Critical Value(s)
• The first or boundary value(s) of the critical
  region(s).

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• Illustration
• Critical Region and Critical Value(s).

Critical Value
Critical Region
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b. Determine whether or not the calculated test
statistic is in the critical region.
Location of z
The calculated value of z, z -1.58, is in
the noncritical region. 5. The Results We
need a decision rule.
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• Decision Rule
• a. If the test statistic falls within the
  critical region, we will reject H0. (The
  critical value is part of the critical region.)
• b. If the test statistic is in the noncritical
  region, we will fail to reject H0.
• a. State the decision about H0.
• Decision Fail to reject H0.
• b. State the conclusion about Ha.
• Conclusion There is not enough evidence at the
  0.05 level of significance to show that the
  mean weight of cereal boxes is less than 24.

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• Note
• 1. The null hypothesis specifies a particular
  value of a population parameter.
• 2. The alternative hypothesis can take three
  forms. Each form dictates a specific location of
  the critical region(s).
• 3. For many hypothesis tests, the sign in the
  alternative hypothesis points in the direction in
  which the critical region is located.
• 4. Significance level a

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• Example The mean water pressure in the main
  water pipe from a town well should be kept at 56
  psi. Anything less and several homes will have
  an insufficient supply, and anything greater
  could burst the pipe. Suppose the water pressure
  is checked at 47 random times. The sample mean
  is 57.1. (Assume s 7.) Is there any evidence
  to suggest the mean water pressure is different
  from 56? Use a 0.01.
• Solution
• 1. The Set-Up
• a. Describe the parameter of concern
• The mean water pressure in the main pipe.
• b. State the null and alternative hypotheses.
• H0 m 56
• Ha m ¹ 56

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• 2. The Hypothesis Test Criteria
• a. Check the assumptions
• A sample of n 47 is large enough for the CLT
  to apply.
• b. Identify the test statistic.
• The test statistic is z.
• c. Determine the level of significance a 0.01
  (given)
• 3. The Sample Evidence
• a. The sample information
• b. Calculate the value of the test statistic

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• 4. The Probability Distribution
• a. Determine the critical regions and the
  critical values.
• b. Determine whether or not the calculated test
  statistic is in the critical region.
• The calculated value of z, z 1.077, is in
  the noncritical region.

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• 5. The Results
• a. State the decision about H0.
• Fail to reject H0.
• c. State the conclusion about Ha.
• There is no evidence to suggest the water
  pressure is different from 56.

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• Example An elementary school principal claims
  students receive no more than 30 minutes of
  homework each night. A random sample of 36
  students showed a sample mean of 36.8 minutes
  spent doing homework (assume s 7.5). Is there
  any evidence to suggest the mean time spent on
  homework is greater than 30 minutes? Use a
  0.05.
• Solution
• 1. The parameter of concern m, the mean time
  spent doing homework each night.
• H0 m 30 ()
• Ha m gt 30

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• 2. The Hypothesis Test Criteria
• a. The sample size is n 36, the CLT applies.
• b. The test statistic is z.
• c. The level of significance is given a 0.01.
• 3. The Sample Evidence

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• 4. The Probability Distribution
• The calculated value of z, z 5.44, is in the
  critical region.

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• 5. The Results
• Decision Reject H0.
• Conclusion There is sufficient evidence at the
  0.01 level of significance to conclude the mean
  time spent on homework by the elementary students
  is more than 30 minutes.
• Note Suppose we took repeated sample of size 36.
  
• What would you expect to happen?