A Statistical hypothesis is a conjecture about a population parameter. This conjecture may or may not be true. Eg.1 If we want to find out whether one method is better than the other, we first make the statement of no difference. This first statement will be; “There is no difference between the two methods.” The second statement will be; “There is difference between the two methods.” Eg.2 In criminal proceedings, the accused is assumed to be innocent unless his guilt is established beyond reasonable doubt. The null hypothesis, symbolized by H0, is a statistical hypothesis that states that there is no difference between a parameter and a specific value or that there is no difference between two parameters.

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BUSINESS STATISTICS FOR
DECISION MAKING
COURSE CODE: MAT504
Hypothesis Testing &
Confidence Interval
ICDE
LECTURE SLIDE 4
BY
S. K. OKRAH (PhD)
0245899021
sokrah@gctu.edu.gh

2. Outline
• Introduction
• Steps in Hypothesis Testing
• Large Sample Mean Test
• Small Sample Mean Test
• Proportion Test

3. Objectives
• Guide students to understand the
definitions used in hypothesis testing.
• State the null and alternative hypotheses.
• Find critical values for the z test.

4. Objectives
• State the five steps used in hypothesis
testing.
• Test means for large samples using the z
test.
• Test means for small samples using the t
test.

5. Steps in Hypothesis Testing
 A Statistical hypothesis is a conjecture about a population parameter. This
conjecture may or may not be true.
• Eg.1 If we want to find out whether one method is better than the other, we
first make the statement of no difference. This first statement will be;
• “There is no difference between the two methods.”
• The second statement will be;
• “There is difference between the two methods.”
• Eg.2 In criminal proceedings, the accused is assumed to be innocent unless
his guilt is established beyond reasonable doubt.
 The null hypothesis, symbolized by H0, is a statistical hypothesis that states
that there is no difference between a parameter and a specific value or that
there is no difference between two parameters.

6. Steps in Hypothesis Testing
 The alternative hypothesis, symbolized by H1, is
a statistical hypothesis that states a specific
difference between a parameter and a specific
value or states that there is a difference between
two parameters.
• Thus, for eg.2,
• H0: The accused is not guilty.
• H1: The accused is guilty.
• For eg.1
• H0: There is no difference between the two methods.
• H1: There is difference between the two methods.

7. Steps in Hypothesis Testing
• In statistical hypothesis testing, one of the
statements is accepted. Note that in concluding
hypothesis testing, we either reject H0 or we fail
to reject H0 (we don’t say that we accept H1
even though rejecting H0 means accepting H1).
• Example
• A business researcher is interested in finding
out whether a new marketing strategy will have
any effects on customers. The researcher is
particularly concerned with the success rate of
the strategy.

8. Steps in Hypothesis Testing - Example
 What are the hypotheses to test whether the success
rate will be different from the mean success rate of
82 percent?
 H0: = 82 H1:  82
 This is a two-tailed test.

9. Steps in Hypothesis Testing - Example
 A chemist invents an additive to increase the
life of an automobile battery. If the mean
lifetime of the battery is 36 months, then his
hypotheses are
 H0:  
 36 H1:   36
 This is a right-tailed test.

10. Steps in Hypothesis Testing - Example
 A contractor wishes to lower heating bills by
using a special type of insulation in houses. If the
average of the monthly heating bills is GHS78, her
hypotheses about heating costs will be
 H0:  78 H1: 78
 This is a left-tailed test.

11. Steps in Hypothesis Testing
 A statistical test uses the data obtained
from a sample to make a decision about
whether or not the null hypothesis should
be rejected.

12. Steps in Hypothesis Testing
 The numerical value obtained from a
statistical test is called the test value.
 In the hypothesis-testing situation,
there are four possible outcomes.

13. Steps in Hypothesis Testing
• In reality, the null hypothesis may or may not be
true, and a decision is made to reject or not to
reject it on the basis of the data obtained from a
sample.
• These decisions may mean two types of error.
• 1. It is possible that H0 may be rejected when in
actual fact H0 is true (i.e. One possible error).
• 2. It is also possible that H0 may not be rejected
when in actual fact H0 is false.
• The first (1) is known as TYPE I ERROR and the
second (2) is known as TYPE II ERROR.

14. Steps in Hypothesis Testing
Error
Type I
Correct
decision
Correct
decision
Error
Type II
Error
Type I
Correct
decision
Correct
decision
Error
Type II
H0 True H0 False
Reject
H0
Do not
reject
H0

15.  DECISION TABLE FOR HYPOTHEIS
 Type I error is very dangerous and thus why
the focus is always on it and it is denoted by α
 A type I error occurs if one rejects the null
hypothesis when it is true.
 A type II error occurs if one does not reject the
null hypothesis when it is false.
Steps in Hypothesis Testing
DECISION H0 is True H0 is False
Reject H0 Type I error (α - Risk) Good decision
Fail to
reject H0
Good decision Type II error (β -
Risk)
.

16.  The level of significance is the maximum
probability of committing a type I error.
This probability is symbolized by  (Greek
letter alpha). That is, P(type I error)=.
 P(type II error) =  (Greek letter beta).
Steps in Hypothesis Testing

17. • Typical significance levels are: 0.10,
0.05, and 0.01.
• For example, when  = 0.10, there is a
10% chance of rejecting a true null
hypothesis.
Steps in Hypothesis Testing

18.  The critical value(s) separates the
critical region from the noncritical
region.
 The symbol for critical value is C.V.
Steps in Hypothesis Testing

19.  The critical or rejection region is the
range of values of the test value that
indicates that there is a significant
difference and that the null hypothesis
should be rejected.
Steps in Hypothesis Testing

20.  The noncritical or nonrejection region is
the range of values of the test value
that indicates that the difference was
probably due to chance and that the
null hypothesis should not be rejected.
Steps in Hypothesis Testing

21.  A one-tailed test (right or left) indicates
that the null hypothesis should be
rejected when the test value is in the
critical region on one side of the mean.
Steps in Hypothesis Testing

22. Large Sample Mean Test
 The z test is a statistical test for the
mean of a population. It can be used
when n  30, or when the population
is normally distributed and  is
known.
 The formula for the z test is given on
the next slide.

23. Large Sample Mean Test
z
X
n
where
X sample mean
hypothesized population mean
population deviation
n sample size











24. • A researcher reports that the average salary of
assistant professors is more than GHS42,000. A
sample of 30 assistant professors has a mean
salary of GHS43,260. At  = 0.05, test the
claim that assistant professors earn more than
GHS42,000 a year. The standard deviation of
the population is GHS5230.
Large Sample Mean Test - Example

25.  Step 1: State the hypotheses and identify the
claim.
 H0: 42,000 H1:  42,000
 Step 2: Find the critical value. Since
 = 0.05 and the test is a right-tailed test, the
critical value is z = +1.65.
 Step 3: Compute the test value.
Large Sample Mean Test - Example

26.  Step 3: z = [43,260 – 42,000]/[5230/30] =
1.32.
 Step 4: Make the decision. Since the test value,
+1.32, is less than the critical value, +1.65, and
not in the critical region, the decision is “Do not
reject the null hypothesis.”
Large Sample Mean Test - Example

27.  Step 5: Summarize the results. There is not
enough evidence to support the claim that
assistant professors earn more on average than
GHS42,000 a year.
 See the next slide for the figure.
Large Sample Mean Test - Example

28. Large Sample Mean Test - Example
1.65
1.32
 = 0.05
Reject

29. • A national magazine claims that the average
university student watches less television than
the general public. The national average is 29.4
hours per week, with a standard deviation of 2
hours. A sample of 30 university students has a
mean of 27 hours. Is there enough evidence to
support the claim at
 = 0.01?
Large Sample Mean Test - Example

30.  Step 1: State the hypotheses and identify the
claim.
 H0: 29.4 H1:  29.4
 Step 2: Find the critical value. Since
 = 0.01 and the test is a left-tailed test, the
critical value is z = –2.33.
 Step 3: Compute the test value.
Large Sample Mean Test - Example

31.  Step 3: z = [27– 29.4]/[2/30] = –6.57.
 Step 4: Make the decision. Since the test value,
–6.57, falls in the critical region, the decision is
to reject the null hypothesis.
Large Sample Mean Test - Example

32.  Step 5: Summarize the results. There is enough
evidence to support the claim that university
students watch less television than the general
public.
 See the next slide for the figure.
Large Sample Mean Test - Example

33. Large Sample Mean Test - Example
-6.57
-2.33
Reject

34. • The Ghana Health Service reports that the
average cost of rehabilitation for stroke victims
is GHS24,672. To see if the average cost of
rehabilitation is different at a large hospital, a
researcher selected a random sample of 35
stroke victims and found that the average cost
of their rehabilitation is GHS25,226.
Large Sample Mean Test - Example

35.  The standard deviation of the population is
GHS3,251. At  = 0.01, can it be concluded
that the average cost at a large hospital is
different from GHS24,672?
 Step 1: State the hypotheses and identify the
claim.
 H0: 24,672 H1:  24,672
Large Sample Mean Test - Example

36.  Step 2: Find the critical values. Since  =
0.01 and the test is a two-tailed test, the critical
values are z = –2.58 and +2.58.
 Step 3: Compute the test value.
 Step 3: z = [25,226 – 24,672]/[3,251/35] =
1.01.
Large Sample Mean Test - Example

37.  Step 4: Make the decision. Do not reject the
null hypothesis, since the test value falls in the
noncritical region.
 Step 5: Summarize the results. There is not
enough evidence to support the claim that the
average cost of rehabilitation at the large
hospital is different from GHS24,672.
Large Sample Mean Test - Example

38. Large Sample Mean Test - Example
-2.58
Reject
2.58

Reject

39. P-Values
• Besides listing an  value, many computer
statistical packages give a P-value for
hypothesis tests. The P-value is the actual
probability of getting the sample mean value
or a more extreme sample mean value in the
direction of the alternative hypothesis (> or <)
if the null hypothesis is true.

40. P-Values
• If the P-value is less than  , the null
hypothesis should be rejected
• And if P-value is greater than or equal to ,

you should fail to reject the null hypothesis.

41.  When the population standard deviation is
unknown and n < 30, the z test is inappropriate
for testing hypotheses involving means.
 The t test is used in this case.
Small Sample Mean Test

42. Small Sample Mean Test - Formula for t test
t
X
s n
where
X sample mean
hypothesized population mean
s sample deviation
n sample size
of freedom n






 


standard
degrees 1

43. • A job placement director claims that the
average starting salary for bankers is different
from GHS24,000. A sample of 10 bankers has a
mean of GHS23,450 and a standard deviation
of GHS400. Is there enough evidence to reject
the director’s claim at  = 0.05?
Small Sample Mean Test - Example

44.  Step 1: State the hypotheses and identify the
claim.
 H0: 24,000 H1:  24,000
 Step 2: Find the critical value. Since
 = 0.05 and the test is a two-tailed test, the
critical values are t = –2.262 and +2.262 with d.f. =
9.
Small Sample Mean Test - Example

45.  Step 3: Compute the test value.
t = [23,450 – 24,000]/[400/] = – 4.35.
 Step 4: Reject the null hypothesis, since
– 4.35 < – 2.262.
 Step 5: There is enough evidence to reject the
claim that the starting salary of nurses is
24,000.
Small Sample Mean Test - Example

46. Small Sample Mean Test - Example
–2.262 2.262


47. • Since the normal distribution can be used to
approximate the binomial distribution when np
5 and nq 5, the standard normal
distribution can be used to test hypotheses for
proportions.
• The formula for the z test for proportions is
given on the next slide.
Proportion Test

48. Formula for the z Test for Proportions
z
X
or z
X np
npq
where
np
npq











49. • An educator estimates that the dropout rate for
students at senior high schools in Ghana is
15%. Last year, 38 students from a random
sample of 200 senior high schools withdrew. At
 = 0.05, is there enough evidence to reject
the educator’s claim?
Proportion Test - Example

50.  Step 1: State the hypotheses and identify the claim.
 H0: p0.15 H1: p 0.15
 Step 2: Find the mean and standard deviation.  =
np = (200)(0.15) = 30 and
 = [
  (200)(0.15)(0.85)] = 5.05.
 Step 3: Find the critical values. Since
 = 0.05 and the test is two-tailed the critical
values are z = 1.96.
Proportion Test - Example

51.  Step 4: Compute the test value. z =
[38 – 30]/[5.05] = 1.58.
 Step 5: Do not reject the null hypothesis,
since the test value falls outside the critical
region.
Proportion Test - Example

52.  Step 6: Summarize the results. There is not
enough evidence to reject the claim that the
dropout rate for students in senior high schools
in Ghana is 15%.
 Note: For one-tailed test for proportions,
follow procedures for the large sample mean
test.
Proportion Test - Example

53. Proportion Test - Example
-1.96 1.96


54. Confidence Intervals and Sample Size

55. Outline
• Introduction
• Confidence Intervals for the Mean [
Known or n 30] and Sample Size
• Confidence Intervals for the Mean [
Unknown and n 30]

56. Outline
• Confidence Intervals and Sample Size for
Proportions
• Confidence Intervals for Variances and
Standard Deviations

57. Objectives
• Find the confidence interval for the
mean when  is known or n 30.
• Determine the minimum sample size for
finding a confidence interval for the
mean.

58. Objectives
• Find the confidence interval for the mean when  is
unknown and n 30.
• Find the confidence interval for a proportion.
• Determine the minimum sample size for finding a
confidence interval for a proportion.
• Find a confidence interval for a variance and a standard
deviation.

59. Confidence Intervals for the Mean ( Known
or n  30) and Sample Size

60.  The estimator must be an unbiased estimator.
That is, the expected value or the mean of the
estimates obtained from samples of a given
size is equal to the parameter being estimated.
Three Properties of a Good Estimator

61.  The estimator must be consistent. For a
consistent estimator, as sample size
increases, the value of the estimator
approaches the value of the parameter
estimated.
Three Properties of a Good Estimator

62.  The estimator must be a relatively
efficient estimator. That is, of all the
statistics that can be used to estimate a
parameter, the relatively efficient
estimator has the smallest variance.
Three Properties of a Good Estimator

63. Confidence Intervals
 An interval estimate of a parameter is an
interval or a range of values used to
estimate the parameter. This estimate
may or may not contain the value of the
parameter being estimated.

64. Confidence Intervals
 A confidence interval is a specific
interval estimate of a parameter
determined by using data obtained from
a sample and the specific confidence
level of the estimate.

65. Confidence Intervals
 The confidence level of an interval
estimate of a parameter is the
probability that the interval estimate will
contain the parameter.

66.  The confidence level is the percentage
equivalent to the decimal value of 1 – .
Formula for the Confidence Interval of the
Mean for a Specific 
X z
n
X z
n






   






 



2 2

67.  The maximum error of estimate is the
maximum difference between the point
estimate of a parameter and the actual
value of the parameter.
Maximum Error of Estimate

68. • The VC of a large university wishes to estimate the
average age of the students presently enrolled.
From past studies, the standard deviation is
known to be 2 years. A sample of 50 students is
selected, and the mean is found to be 23.2 years.
Find the 95% confidence interval of the
population mean.
Confidence Intervals - Example

69. Since the confidence
is desired z Hence
substituting in the formula
X z
n
X z
n
onegets
, ,
– +
2
95%
196
2 2
interval
a
a a
s
m
s
=
æ
è
ç
ö
ø
÷ < < æ
è
ç
ö
ø
÷
. .
Confidence Intervals - Example

70. Confidence Intervals - Example

71. • A certain medication is known to increase
the pulse rate of its users. The standard
deviation of the pulse rate is known to be 5
beats per minute. A sample of 30 users had
an average pulse rate of 104 beats per
minute. Find the 99% confidence interval
of the true mean.
Confidence Intervals - Example

72. Confidence Intervals - Example

73. Confidence Intervals - Example

74. Formula for the Minimum Sample Size Needed
for an Interval Estimate of the Population Mean
.
,
.
n
z
E
where E is the error
of estimate
If necessary round the answer up
to obtain a whole number










2
2
maximum

75. • The VC of a university wants to estimate the
average age of the students at their university.
How large a sample is necessary? He decides the
estimate should be accurate within 1 year and be
99% confident. From a previous study, the
standard deviation of the ages is known to be 3
years.
Minimum Sample Size Needed for an Interval
Estimate of the Population Mean - Example

76. Minimum Sample Size Needed for an Interval
Estimate of the Population Mean - Example

77. .
,
.
n pq
z
E
where E is the error
of estimate
If necessary round the answer up
to obtain a whole number







   2
2
maximum
Sample Size Needed for Interval Estimate of a
Population Proportion

78. • A researcher wishes to estimate, with 95%
confidence, the number of people who own a
home computer. A previous study shows that
40% of those interviewed had a computer at
home. The researcher wishes to be accurate
within 2% of the true proportion. Find the
minimum sample size necessary.
Sample Size Needed for Interval Estimate of a
Population Proportion - Example

79. Sample Size Needed for Interval Estimate of a
Population Proportion - Example

80. • Ten randomly selected automobiles were
stopped, and the tread depth of the right
front tire was measured. The mean was 0.32
inch, and the standard deviation was 0.08
inch. Find the 95% confidence interval of the
mean depth. Assume that the variable is
approximately normally distributed.
Confidence Interval for the Mean
( Unknown and n < 30) - Example

81. • Since  is unknown and s must replace it,
the t distribution must be used with
 = 0.05. Hence, with 9 degrees of freedom,
t/2 = 2.262 (see the t-distribution Table).
• From the next slide, we can be 95% confident
that the population mean is between 0.26
and 0.38.
Confidence Interval for the Mean
( Unknown and n < 30) - Example

82. Confidence Interval for the Mean
( Unknown and n < 30) - Example

83. Confidence Intervals and Sample Size for
Proportions

84. Confidence Intervals and Sample Size for
Proportions - Example
p̂ q̂

85. Confidence Intervals and Sample Size for
Proportions - Example

86. Formula for a Specific Confidence Interval for a
Proportion

87. Specific Confidence Interval for a Proportion -
Example

88. Specific Confidence Interval for a
Proportion - Example

89. .
,
.
n pq
z
E
where E is the error
of estimate
If necessary round the answer up
to obtain a whole number







   2
2
maximum
Sample Size Needed for Interval Estimate of a
Population Proportion

90. • A researcher wishes to estimate, with 95%
confidence, the number of people who own a
home computer. A previous study shows that
40% of those interviewed had a computer at
home. The researcher wishes to be accurate
within 2% of the true proportion. Find the
minimum sample size necessary.
Sample Size Needed for Interval Estimate of a
Population Proportion - Example

91. Sample Size Needed for Interval Estimate of a
Population Proportion - Example