Review

1
Review 2

• Chapter 9
• Chapter 10
• Chapter 11 and 12

2
Chapter 9Sampling Distributions

• A statistic is a random variable describing a
  characteristic of a random samples.
• Sample mean
• Sample variance
• We use statistic values in inferential statistics
  (make inference about population characteristics
  from sample characteristics).
• Statistics have distributions of their own.

3
Chapter 9 The Central Limit Theorem

• The distribution of the sample mean is normal if
  the parent distribution is normal.
• The distribution of the sample mean approaches
  the normal distribution for sufficiently large
  samples (n ³ 30), even if the parent
  distribution is not normal.
• The parameters of the sample distribution of the
  mean are
• Mean
• Standard deviation
• (Assumption The population is sufficiently
  large. No correction is needed in the
  calculation of the variance).

4
Chapter 9 The Central Limit Theorem

• Problem 1 (Using Excel) Given a normal
  population whose mean is 50 and whose standard
  deviation is 5,
• Question 1 Find the probability that a random
  sample of 4 has a mean between 49 and 52
• Answer

-.4
.8
5
Chapter 9The Central Limit Theorem
Normal table

• Problem 1 (Using the table) Given a normal
  population whose mean is 50 and whose standard
  deviation is 5,
• Question 1 Find the probability that a random
  sample of 4 has a mean between 49 and 52
• Answer

-.4
.8
6
Chapter 9The Central Limit Theorem
Normal table

• Problem 1
• Question 2 Find the probability that a random
  sample of 16 has a mean between 49 and 52.
• Answer

7
Chapter 9 The Central Limit Theorem
Normal table

• Problem 2 The amount of time per day spent by
  adults watching TV is normally distributed with
  m6 and s1.5 hours.
• Question 1 What is the probability that a
  randomly selected adult watches TV for more
  than 7 hours a day?
• Answer

• Question 2 What is the probability that 5 adults
  watch TV on the average 7 or more hours?Answer

8
Chapter 9 The Central Limit Theorem
Normal table

• Problem 2
• Question 3 What is the probability that the
  total time of watching TV of the five adults will
  not exceed 28 hours?
• Answer
• Question 4 What total TV watching time is
  exceeded by only 3 of the population for samples
  of 5 adults?

Comments 1.Excel returns X for agiven left hand
tail probability 2. .670822 1.5/5.5
9
Chapter 9 The Central Limit Theorem
Normal table

• Problem 3
• Assume that the monthly rents paid by students
  in a particular town is 350 with a standard
  deviation of 40. A random sample of 100 students
  who rented apartments was taken.
• Question1 What is the probability that the
  sample mean of the monthly rent exceeds 355?

10
Chapter 9 The Central Limit Theorem
Normal table

• Problem 3 - continued
• Question2 What is the probability that the
  total revenue from renting 10 randomly selected
  apartments falls between 3300 and 3700 dollars?

11
Chapter 9 The Central Limit Theorem
Normal table

• Problem 3 - continued
• Question3 Lets assume the population mean was
  unknown, but the standard deviation was known to
  be 40. A sample of 100 rentals was selected in
  order to estimate the mean monthly rent paid by
  the whole student population. What is the
  probability that the sample mean differ from the
  actual mean by more than 5? How about more than
  10?

12
Chapter 9 The Central Limit Theorem

• Problem 3
• continued

13
Chapter 9Sampling distribution of the sample
proportion

• In a sample of size n, if np gt 5 and n(1-p) gt 5,
  then the sample proportion p x/n is
  approximately normally distributed with the
  following parameters

(Assumption The population is sufficiently
large. No correction is needed in the
calculation of the variance).
14
Sampling distribution of the sample proportion

• Problem 4
• A commercial of a household appliances
  manufacturer claims that less than 5 of all of
  its products require a service call in the first
  year.
• A survey of 400 households that recently
  purchased the manufacturer products was conducted
  to check the claim.

15
Sampling distribution of the sample proportion
Normal table

• Problem 4 - Continued Assuming the
  manufacturer is right, what is the probability
  that more than 10 of the surveyed households
  require a service call within the first year?

If indeed 10 of the sampled households reported
a call for service within the first year, what
does ittell you about the the manufacturer
claim?
16
Sampling Distribution of the Difference Between
two Means

• If two independent variables are normally
  distributed with means and variances m1, s21,
  and m2, s22 respectively, then x1 x2 is also
  normally distributed with

17
Sampling Distribution of the Difference Between
two Means

• When at least one of the populations is not
  normally distributed but the samples sizes are
  both at least 30, x1 x2 is approximately
  normally distributed, with a mean and a variance
  as indicated above.

18
Sampling Distribution of the Difference Between
two Means

• Example A national TV telethon committee is
  interested in determining whether donations made
  by males are on the average larger than those
  made by females by 4. Two samples of 25 males
  and 25 females were selected, and the donations
  made recorded. If the standard deviations of the
  male and female populations are 2.4 and 1.8
  respectively, what is the probability that sample
  mean of the male donations exceeds the sample
  mean of the female donations by at least 5?
  Assume donations for the two populations are
  normally distributed.

19
Sampling Distribution of the Difference Between
two Means

• Solution

For males For females
20
Chapter 10Introduction to Estimation

• A populations parameter can be estimated by a
  point estimator and by an interval estimator.
• A confidence interval with 1-a confidence level
  is an interval estimator that covers the
  estimated parameters (1-a) of the time.
• Confidence intervals are constructed using
  sampling distributions.

21
Confidence interval of the mean Known Variance

• We use the central limit theorem to build the
  following confidence interval

22
Confidence interval of the mean Known Variance

• Problem 5 How many classes university students
  miss each semester? A survey of 100 students was
  conducted. (See Data next)
• Assuming the standard deviation of the number of
  classes missed is 2.2, estimate the mean number
  of classes missed per student. Use 99 confidence
  level.

23
Confidence interval of the mean Known Variance
Data

• Solution 10.21 2.575
  10.21 .57

1- a .99 a .01 a/2 .005 Za/2 Z.005 2.575
LCL 9.64, UCL 10.78
You can used Data Analysis Plus gt Z-Estimate Mean
24
Confidence interval of the mean Known Variance
Data

• Solution (using Data Analysis Plus)
• Shade the data set (you may include the title
  label)
• Select Data Analysis Plus, then Z-Estimate
  Mean
• Type in the sigma (2.2), check Labels (if
  appropriate), type in alpha (.01), click OK.

25
Selecting the sample size

• The shorter the confidence interval, the more
  accurate the estimate.
• We can, therefore, limit the width of the
  interval to 2W, and get
• From here we have

W is called Margin of error, or Bound on the
error estimate
26
Selecting the sample size

• Problem 6An operation manager wants to estimate
  the average amount of time needed by a worker to
  assemble a new electronic component.
• Sigma is known to be 6 minutes.
• The required estimate accuracy is within 20
  seconds.
• The confidence level is 90 95.
• Find the sample size.

27
Selecting the sample size

• Solution
• s 6 min W 20 sec 1/3 min
• 1 - a .90 Za/2 Z.05 1.645
• 1-a .95, Za/2 Z.025 1.96

28
Chapter 11Hypotheses tests

• In hypothesis tests we hypothesize on a value of
  a population parameter, and test to see if there
  is sufficient evidence to support our belief.
• The structure of hypotheses test
• Formulate two hypotheses.
• H0 The one we try to reject in favor of
• H1 The alternative hypothesis, the one we try to
  prove.
• Define a significance level a.

29
Hypotheses tests

• The significance level is the probability of
  erroneously reject the null hypothesis.
• a P(reject H0 when H0 is true)
• Sample from the population and calculate a
  statistic that provides an indication whether or
  not the parameter value under H1 is more likely
  to be true.
• We shall test the population mean assuming the
  standard deviation is known.

30
Hypotheses tests of the Mean Known Variance

• Problem 7 A machine is set so that the average
  diameter of ball bearings it produces is .50
  inch. In a sample of 100 ball bearings the mean
  diameter was .51 inch. Assuming the standard
  deviation is .05 inch, can we conclude at 5
  significance level that the mean diameter is not
  .50 inch.

31
Hypotheses tests of the Mean Known Variance

• SolutionThe population studied is the
  ball-bearing diameters.
• We hypothesize on the population mean.
• A good point estimator for the population mean is
  the sample mean.
• We use the distribution of the sample mean to
  build a sample statistic to test whether m .50
  inch.

32
Hypotheses tests of the Mean Known Variance

• Solution (A Two Tail rejection region)
• Define the hypotheses
• H0 m .50
• H1 m .50

The probability of conducting atype one error
33
Hypotheses tests of the Mean Known Variance
Solution - A Two Tail rejection region
Critical Z
Z.025 1.96 (obtained from the Z-table) Build a
rejection region Zsamplegt Za/2, or
Zsamplelt-Za/2
1.96
-1.96
Calculate the value of the sample Z statistic
and compare it to the critical value
Since 2 gt 1.96, there is sufficient evidence to
rejectH0 in favor of H1 at 5 significance
level.
34
Hypotheses tests of the Mean Known Variance
Solution - A Two Tail rejection region

• We can perform the test in terms of the mean
  value.
• Let us find the critical mean values for
  rejection
• XL2m0 Z.025 .501.96(.05)/(100)1/2
  .5098
• XL1m0 - Z.025 .50
  -1.96(.05)/(100)1/2.402

Since.51 gt .5098, there is sufficient evidence to
reject the null hypothesis at 5 significance
level.
35
Hypotheses tests of the Mean Known Variance

• Calculate the p value of this test
• Solutionp-value P(Z gt Zsample) P(Z lt
  -Zsample) P(Z gt 2) P(Z lt -2) 2P(Z gt 2)
  21 - .9772 .0456
• Since .0456 lt .05, H0 is rejected.

36
Hypotheses tests of the Mean Known Variance

• Problem 8
• The average annual return on investment for
  American banks was found to be 10.2 with
  standard deviation of 0.8.
• It is believed that banks that exercise
  comprehensive planning do better.
• A sample of 26 banks that exercise comprehensive
  training provide the following result Mean
  return 10.5
• Can we infer that the belief about bank
  performance is supported at 10 significance
  level by this sample result?

37
Hypotheses tests of the Mean Known Variance
Data

• Solution (A right Hand Tail Rejection
  region)The population tested is the annual rate
  of return.
• H0 m 10.2
• H1 m gt 10.2
• Let us perform the test with the standardized
  rejection region approach Zsample gt Z.10
  (Right hand tail rejection region) Z.10 1.28.
  Reject H0 if Zsample gt 1.28

38
Hypotheses tests of the Mean Known Variance

• Conclusion
• At 10 significance level there is sufficient
  evidence in the data to reject H0 in favor of H1,
  since the sample statistic falls inside the
  rejection region.
• Interpretation
• If we are willing to accept 10 chance of making
  the wrong conclusion, we can conclude banks
  conducting comprehensive training perform better
  than banks who do not.

39
Hypotheses tests of the Mean Known Variance
Data

• Let us perform the test with the p-value method
• P(X gt 10.5 given that m 10.2) P(Z gt (10.5
  10.2)/.8/(26)1/2 P(Z gt 1.91) .5 - .4719
  .0281
• Since .0281 lt .10 we reject the null hypothesis
  at 10 significance level.

40
Hypotheses tests of the Mean Known Variance

• Note the equivalence between the standardized
  method or the rejection region method and the
  p-value method.
• P(ZgtZ.10) .10Z10 1.28

The statement p-value is smallerthan alpha, is
equivalent to the statement the test statistic
fallsin the rejection region
1.91
1.28
41
Hypotheses tests of the Mean Known Variance

• Problem 9
• In the midst of labor-management negotiations,
  the president of a company argues that the
  companys blue collar workers, who are paid an
  average of 30K a year, are well-paid because the
  mean annual pay for blue-collar workers in the
  country is less than 30K.
• This figure is disputed by the union. To test the
  presidents belief an arbitrator draws a random
  sample of 350 blue-collar workers from across the
  country and their income recorded (see file
  Salaries).
• If the arbitrator assumes that income is normally
  distributed with a standard deviation of 8,000,
  can it be inferred at 5 significance level that
  the companys president is correct?

42
Hypotheses tests of the Mean Known Variance
Data

• Solution (A left Hand Tail Rejection Region)The
  population tested is the ann. Salary
• H0 m 30KH1 m lt 30K
• Left hand Tail Rejection region Z lt -Z.05 or Z lt
  -1.645ZSample (29,119.5-30,000)/(8,000/350.5)
  -2.059Since 2.059 lt -1.645 there is sufficient
  evidence to infer that on the average blue collar
  workers income is lower than 30K at 5
  significance level.

43
Hypotheses tests of the Mean Known Variance

• Calculate the p-value of this test
• Solutionp-value P(Z lt Zsample) P(Z lt -2.059)

44
Type II Error

• Problem 7a Calculate b for the two-tail
  hypotheses test performed in problem 7, when the
  actual mean diameter is .515 inch.
• Solution
• The rejection region in terms of the critical
  values of the sample mean was found before XL1
  .402 XL2 .5098.
• b P(Do not reject H0 when H1 is true)
  P(.402 lt lt .5098 when m .515)
  P(.402-.515)/.05/(100).5 lt Z lt
  (.5098-.515)/.05/(100).5 P(-22.6 lt Z lt -1.04)
  P(1.04 lt Z lt 22.6)
• 1 - .8508 .1492
• This large probability may be reduced by taking
  larger samples

H0 m .500H1 m .515
P(Zlt22.6) P(Zlt1.04) 1-P(Zlt1.04)
45
Ch 12 Inference when the Variance is Unknown

• Generally, the variance may be unknown
• In this case we change the test statistic from
  Z to t, when testing the population mean.
• To test the population proportion well use the
  normal distribution (under certain conditions).

46
Testing the mean unknown variance

• Replace the statistic Z with t
• The original distribution must be normal (or at
  least mound shaped).

47
Testing the mean unknown variance

• Problem 10
• A federal agency inspects packages to determine
  if the contents is at least as large as that
  advertised.
• A random sample of (i)5, (ii)50 containers whose
  packaging states that the weight was 8.04 ounces
  was drawn. (data is provided later)
• From the sample results
• Can we conclude that the average weight does not
  meet the weight stated? (use a .05).
• Estimate the mean weight of all containers with
  99 confidence
• What assumption must be met?

48
Testing the mean unknown variance

• Solution
• We hypothesize on the mean weight.
• H0 m 8.04
• H1 m lt 8.04
• (i) n5. For small samples let us solve
  manuallyAssume the sample was 8.07, 8.03,
  7.99, 7.95, 7.94
• The rejection region t lt -ta, n-1 -t.05,5-1
  -2.132The tsample ?
• Mean (8.077.94)/5 7.996Std.
  Dev.(8.07-7.996)2(7.94- 7.996)2/41/2
  0.054

-2.132
49
Testing the mean unknown variance

• The tsample is calculated as follows
• Since -1.32 gt -2.132 the sample statistic does
  not fall in the rejection region. There is
  insufficient evidence to conclude that the mean
  weight is smaller than 8, at 5 significance
  level.

-.165
-2.132
50
Testing the mean unknown variance

• (ii) n50. To calculate the sample statistics we
  use Excel, Descriptive statistics from the
  ToolsgtData analysis menu. From the sample we
  obtainMean 8.02 Std. Dev. .04
• The confidence interval is calculated by
  
• 8.02 2.678
  8.02 .015

LCL 8.005, UCL 8.35
51
Testing the mean unknown variance
Data

• Comments
• Check whether it appears that the distribution is
  normal

52
Using Excel
Data

• To obtain an exact value for t use the TINV
  function
• The exact value

Degrees of freedom
TINV(0.01,49)
.01 is the two tail probability .0052
2.6799535
53
Testing the mean unknown variance

• Problem 11
• Engineers in charge of the production of car
  seats are concerned about the compliance of the
  springs used with design specifications.
• Springs are designed to be 500mm long.
• Springs too long or too short must be reworked.
• A standard deviation of 2mm in springs length
  will result in an acceptable number of reworked
  springs.
• A sample of 100 springs was taken and measured.

54
Testing the mean unknown variance
Data

• Problem continued
• Can we infer at 10 significance level that the
  mean spring length is not 500mm?

SolutionH0 m 500 Since the standard
deviation is unknown H1 m ¹ 500 We need to
run a t-test, assuming the
spring length is normally distributed.
Rejection region t lt -ta/2 or t gt ta/2with d.f.
99
t lt -1.6604 ort gt 1.6604
55
Inference about a population proportion

• The test and the confidence interval are based on
  the approximated normal distribution of the
  sample proportion, if npgt5 and n(1-p)gt5.
• For the confidence interval of p we have
• where p x/n
• For the hypotheses test, we use a Z test.

56
Inference about a population proportion

• Problem 12 (problem 11 continued). The engineers
  were interested in the percentage of springs that
  are the correct length. They marked each spring
  in the sample as
• Correct 1
• Too long 2
• Too short 3

Can we infer that less than 90 of the springs
are the correct length, at 10 sig.
level?
57
Inference about a population proportion
Data

• Problem 12 - Solution
• H0 p .9H1 p lt .9
• Rejection regionZ lt -Za, or Z lt -1.28

ConclusionSince 1.33 lt -1.28 we can infer
that less than 90 of the springs do not need
reworking.
58
Inference about a population proportion
Data

• Problem 12 solution continued
• Let us estimate the proportion of good springs at
  99 confidence level.

59
Inference about a population proportion

• Problem 12 solution continued
• Find the sample size if the proportion of good
  springs is to be estimated to within .035.
  Consider the given sample an initial sample.

60
Inference about a population proportion

• Problem 13
• A consumer protection group runs a survey of 400
  dentists to check a claim that more than 4 out of
  5 dentists recommend ingredients included in a
  certain toothpaste.
• The survey results are as follows 71 No 329
  Yes
• At 5 significance level, can the consumer group
  infer that the claim is true?

61
Inference about a population proportion

• Problem 13 - Solution
• The two hypotheses are
• H0 p .8
• H1 p gt .8
• Z.05 1.645
• Conclusion Since 1.125 lt 1.645 the consumer
  group cannot confirm the claim at 5 significance
  level.

The rejection region Z gt Za
62
Summary Example

• An automotive expert claims that the large number
  of self-serve gas stations has resulted in poor
  automobile maintenance, and that the average tire
  pressure is more than 4.5 psi below its
  manufacturer specifications.
• A random sample of 50 tires revealed the results
  stored in the file TirePressure.
• Assume the tire pressure is normally distributed
  with s 1.5 psi, and answer the following
  questions

63
Summary Example
Tire Pressure

• At 10 significance level can we infer that the
  expert is correct? What is the p value?

• Solution
• The HypothesesH0 m 4.5H1 m gt 4.5 The
  rejection region Z gt Z.10 or Z gt 1.28.From the
  data we have mean 5.04, soZ(5.04
  4.5)/(1.5/50.5) 2.545
• Since 2.545 gt 1.28, there is sufficient evidence
  to infer that the expert is correct.

The p value P(Sample Mean gt 5.04 when m
4.5)P(Z gt 2.545) 1- .9945 .0055
64
Summary Example

• Find the probability of making a type II error
  when the actual tire under-inflation is 5 psi on
  the average.
• SolutionThe Rejection Region in terms of the
  sample means is found firstZL 1.28 (XL
  4.5)/(1.5/50.5). XL 4.5 1.28(1.5/50.5)
  4.77. So, the Rejection Region is Sample mean
  gt 4.77. b P(accept H0 when H1 is true)
  P(sample mean does not fall in the RR, when m
  5) P( lt 4.77 when m 5) P(Z lt
  (4.77-5)/(1.5/50.5)) P(Z lt -1.08)
• From Excel NORMSDIST(-1.077) .1407

65
Inference about the population Variance

• The following statistic is c2 (Chi squared)
  distributed with n-1 degrees of freedom
• We use this relationship to test and estimate the
  variance.

66
Inference about the population Variance

• The Hypotheses tested are
• The rejection region is

67
Testing the Variance

• Problem 15
• Engineers in charge of the production of car
  seats are concerned about the compliance of the
  springs used with design specifications.
• Springs are designed to be 500mm long.
• Springs too long or too short must be reworked.
• A standard deviation of 2mm in springs length
  will result in an acceptable number of reworked
  springs.
• A sample of 100 springs was taken and measured.

68
Testing the Variance
Data

• Problem 15 - continued Can we infer at 10
  significance level that the number of springs
  requiring reworking is unacceptably large?

H0 s2 4 H1 s2 gt 4
The number of springs requiring reworkingdepends
on the standard deviation, or the variance.
Rejection regionc2Sample gt c2ad.f. 99
c2Sample gt 117.4069
69
Testing the Variance

• Problem 15 - conclusion Since 161.25 gt 117.4069,
  we can infer at 10 significance level that the
  standard deviation is greater than 2, thus the
  number of springs that require reworking is
  unacceptably large.

70
Testing the Variance

• Problem 16
• A random sample of 100 observations was taken
  from a normal population. The sample variance
  was 29.76.
• Can we infer at 2.5 significance level that the
  population variance DOES NOT exceeds 30?
• Estimate the population variance with 90
  confidence.

71
Testing the Variance

• Problem 16 Solution
• H0s2 30
• H1s2 lt 30
• c2
  98.21

Rejection region c2 lt c21-a, n-1 c2 lt 73.36
!
72
Testing the Variance

• Problem 16 - conclusion Since 98.208 gt 73.36 we
  conclude that there is insufficient evidence at
  2.5 significance level to infer that the
  variance is smaller than 30.

73
Using Excel

• We can get an exact value of the probability
  P(c2d.f.gt c2) ? for a given c2 and known d.f.,
  and then determine the p-value.
• Use the CHIDIST function For example
  .50359
• That is P(c299gt 98.208) .50359
• In our example we had a left hand tail rejection
  region, and therefore the p-value is P(c299 lt
  98.208) 1 - .50359 .49641gt .025

CHIDIST(c2,d.f.)
CHIDIST(98.208,99)
74
Using Excel

• We can get the exact c2 value for which
  P(c2d.f.gt c2) a, for any given probability a
  and known d.f., then define the rejection region
• Use the CHIINV functionFor example
  CHIINV(.975,99) 73.36
• That is P(c299 gt ?) .975. c2 73.36The
  rejection region is c2 lt 73.36.

CHIINV(a,d.f.)