1. Statistical Inferences
Introduction
The field of statistical inference consist of those methods used to make decisions or to draw conclusions about a population. These methods utilize the information contained in a sample from the population in drawing conclusions. Statistical Inference may be divided into two major areas: parameter estimation and hypothesis testing.

2. Confidence Interval
Interval Estimate: In interval estimation, an interval is constructed around the point estimate and it is stated that this interval is likely to contain the corresponding population parameter. Confidence Level and Confidence Interval: Each interval is constructed with regard to a given confidence level and is called a confidence interval. The confidence level associated with a confidence interval states how much confidence we have that this interval contains the true population parameter. The confidence level is denoted by .

3. Confidence Intervals Confidence Intervals Population Mean Population
Proportion
σ Known
σ Unknown
EQT 373

4. Confidence Interval Estimates for Population Mean

6. Example 1
solution 6

8. Example 2 : A publishing company has just published a new textbook
Example 2 : A publishing company has just published a new textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 36 comparable textbooks and collected the information on their prices. this information produced a mean price RM for this sample. It is known that the standard deviation of the prices of all such textbooks is RM4.50. Construct a 90% confidence interval for the mean price of all such college textbooks.

9. solution

14. Example 3 : The scientist wondered whether there was a difference in the average daily intakes of dairy products between men and women. He took a sample of n =50 adult women and recorded their daily intakes of dairy products in grams per day. He did the same for adult men. A summary of his sample results is listed below. Construct a 95% confidence interval for the difference in the average daily intakes of daily products for men and women. Can you conclude that there is a difference in the average daily intakes of daily products for men and women?
Men
Women
Sample size 50
Sample mean
780 grams per day
762 grams per day
Sample standard deviation 35 30

15. Solution

16. Confidence Interval Estimates for Population Proportion

17. Example 4 According to the analysis of Women Magazine in June 2005, “Stress has become a common part of everyday life among working women in Malaysia. The demands of work, family and home place an increasing burden on average Malaysian women”. According to this poll, 40% of working women included in the survey indicated that they had a little amount of time to relax. The poll was based on a randomly selected of 1502 working women aged 30 and above. Construct a 95% confidence interval for the corresponding population proportion.

18. Solution

20. Example 5: A researcher wanted to estimate the difference between the percentages of users of two toothpastes who will never switch to another toothpaste. In a sample of 500 users of Toothpaste A taken by this researcher, 100 said that the will never switch to another toothpaste. In another sample of 400 users of Toothpaste B taken by the same researcher, 68 said that they will never switch to another toothpaste. Construct a 97% confidence interval for the difference between the proportions of all users of the two toothpastes who will never switch.

21. Solutions Toothpaste A : n1 = 500 and x1 = 100 Toothpaste B : n2 = 400 and x2 = 68 The sample proportions are calculated; Thus, with 97% confidence we can state that the difference between the two population proportions is between and

22. Determining Sample Size
For the
Mean
For the
Proportion

23. Error of Estimation and Determining Sample size
The required sample size can be found to reach a desired margin of error (e) with a specified level of confidence (1 - ). The margin of error is also called error of estimation
Estimating sample size of the Population Mean):

24. Determining Sample Size for population mean problems
For the
Mean
Sampling error (margin of error)

25. So the required sample size is n = 220
Example 6
If  = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence
So the required sample size is n = 220
(Always round up)

26. Example 7: sample of size n=150 to estimate the average mechanical
A team of efficiency experts intends to use the mean of a random
sample of size n=150 to estimate the average mechanical
aptitude of assembly-line workers in a large industry (as
measured by a certain standardized test). If, based on
experience, the efficiency experts can assume that
for such data, what can they assert with probability 0.99
about the maximum error of their estimate?

27. Solutions

28. Estimating sample size of the Population Proportion):

29. Determining Sample Size for population proportion problems
For the
Proportion
Now solve for n to get

30. Example 8
How large a sample would be necessary to estimate the true proportion defective in a large population within ±3%, with 95% confidence? (Assume a sample yields p = 0.12) For 95% confidence, we have Zα/2 = 1.96 e = 0.03; p = 0.12
Solution:
So use n = 451

31. Example 9: A study is made to determine the proportion of voters in a sizable community who favor the construction of a nuclear power plant. If 140 of 400 voters selected at random favor the project and we use as an estimate of the actual proportion of all voters in the community who favor the project, what can we say with 99% confidence about the maximum error?

32. Solution

33. Example 10: How large a sample required if we want to be 95% confident
that the error in using to estimate p is less than 0.05? If
, find the required sample size.

34. Solution

35. Hypothesis Testing
Hypothesis and Test Procedures A statistical test of hypothesis consist of : 1. The Null hypothesis, 2. The Alternative hypothesis, 3. The test statistic and its p-value 4. The rejection region 5. The conclusion

36. Definition :
Hypothesis testing can be used to determine whether a statement
about the value of a population parameter should or should not
be rejected.
Null hypothesis, H0 : A null hypothesis is a claim (or statement)
about a population parameter that is assumed to be true.
(the null hypothesis is either rejected or fails to be rejected.)
Alternative hypothesis, H1 : An alternative hypothesis is a claim
about a population parameter that will be true if the null
hypothesis is false.
Test Statistic is a function of the sample data on which the
decision is to be based.
p-value is the probability calculated using the test statistic. The
smaller the p-value, the more contradictory is the data to

37. Hypothesis Testing
Researchers are interested in answering many types of questions. For example,
Is the earth warming up?
Does a new medication lower blood pressure?
Does the public prefer a certain color in a new fashion line?
Is a new teaching technique better than a traditional one?
Do seat belts reduce the severity of injuries?
These types of questions can be addressed through statistical hypothesis testing, which is a decision-making process for evaluating claims about a population.

38. Hypothesis Testing 1. The traditional method 2. The P-value method
Three methods used to test hypotheses:
1. The traditional method
2. The P-value method
3. The confidence interval method

39. Claim
When a researcher conducts a study, he or she is generally looking for evidence to support a claim. Therefore, the claim should be stated as the alternative hypothesis, or research hypothesis.
A claim, though, can be stated as either the null hypothesis or the alternative hypothesis; however, the statistical evidence can only support the claim if it is the alternative hypothesis. Statistical evidence can be used to reject the claim if the claim is the null hypothesis.
These facts are important when you are stating the conclusion of a statistical study.

40. Hypothesis Testing
After stating the hypotheses, the researcher’s next step is to design the study. The researcher selects the correct statistical test, chooses an appropriate level of significance, and formulates a plan for conducting the study.
A statistical test uses the data obtained from a sample to make a decision about whether the null hypothesis should be rejected.
The numerical value obtained from a statistical test is called the test value.
In the hypothesis-testing situation, there are four possible outcomes.

41. 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.
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.

42. Hypothesis Testing

43. Hypothesis Testing
The level of significance is the maximum probability of committing a type I error. This probability is symbolized by a (alpha). That is,
P(type I error) = a.
Likewise,
P(type II error) = b (beta).

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

45. Hypothesis Testing
The critical value, C.V., separates the critical region from the noncritical region.
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.
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.

46. Hypothesis Testing
Finding the Critical Value for α = 0.01 (Right-Tailed Test)
z = 2.33 for α = 0.01

47. Hypothesis Testing
Finding the Critical Value for α = 0.01 (Left-Tailed Test) z
Because of symmetry,
z = for α = 0.01

48. Hypothesis Testing
Finding the Critical Value for α = 0.01 (Two-Tailed Test)
z = ±2.58

49. Finding the Critical Values for Specific α Values, Using Table 5
Procedure Table
Finding the Critical Values for Specific α Values, Using Table 5
Step 1 Draw the figure and indicate the appropriate area.
a. If the test is right-tailed, the critical region, with an area equal to α, will be on the right side of the mean.
b. If the test is left-tailed, the critical region, with an area equal to α, will be on the left side of the mean.
c. If the test is two-tailed, α must be divided by 2; one-half of the area will be to the right of the mean, and one-half will be to the left of the mean.

50. Procedure Table
Step 2 Find the z value in Table 5.
a. For a right-tailed test, use the z value that corresponds to the area equivalent to α in Table 5.
b. For a left-tailed test, use the z value that corresponds to the area equivalent to α. It will be negative.
c. For a two-tailed test, use the z value that corresponds to α / 2 for the right value. It will be positive. For the left value, use the z value that corresponds to the area equivalent to α / 2. It will be negative.

51. Example: Using Table 5
Using Table 5, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region.
a. A left-tailed test with α = 0.10.
Step 1 Draw the figure and indicate the appropriate area.
Step 2 The z value is

52. Example: Using Table 5
Using Table 5, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region.
c. A right-tailed test with α =
Step 1 Draw the figure and indicate the appropriate area.
Step 2 The z value is

53. Example: Using Table 5
Using Table 5, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region.
b. A two-tailed test with α = 0.02.
Step 1 Draw the figure with areas α /2 = 0.02/2 = 0.01.
Step 2 The z values are and 2.33.

54. Solving Hypothesis-Testing Problems
Procedure Table
Solving Hypothesis-Testing Problems
(Traditional Method)
Step 1 State the hypotheses and identify the claim.
Step 2 Find the critical value(s) from the appropriate table 5.
Step 3 Compute the test value.
Step 4 Make the decision to reject or not reject the null hypothesis.
Step 5 Summarize the results.

55. Developing Null and Alternative Hypothesis
It is not always obvious how the null and alternative hypothesis should be formulated.
When formulating the null and alternative hypothesis, the
nature or purpose of the test must also be taken into
account. We will examine:
The claim or assertion leading to the test.
The null hypothesis to be evaluated.
The alternative hypothesis.
Whether the test will be two-tail or one-tail.
A visual representation of the test itself.
In some cases it is easier to identify the alternative hypothesis first. In other cases the null is easier.

56. Alternative Hypothesis as a Research Hypothesis
Many applications of hypothesis testing involve
an attempt to gather evidence in support of a
research hypothesis.
In such cases, it is often best to begin with the
alternative hypothesis and make it the conclusion
that the researcher hopes to support.
The conclusion that the research hypothesis is true
is made if the sample data provide sufficient
evidence to show that the null hypothesis can be
rejected.

57. Example: A new drug is developed with the goal
of lowering blood pressure more than the existing drug.
Alternative Hypothesis:
The new drug lowers blood pressure more than
the existing drug.
Null Hypothesis:
The new drug does not lower blood pressure more
than the existing drug.

58. Null Hypothesis as an Assumption to be Challenged
We might begin with a belief or assumption that
a statement about the value of a population
parameter is true.
We then using a hypothesis test to challenge the
assumption and determine if there is statistical
evidence to conclude that the assumption is
incorrect.
In these situations, it is helpful to develop the null
hypothesis first.

59. Example: The label on a soft drink bottle states
that it contains at least 67.6 fluid ounces.
Null Hypothesis:
The label is correct. µ > 67.6 ounces.
Alternative Hypothesis:
The label is incorrect. µ < 67.6 ounces.

60. Example: Average tire life is 35000 miles.
Null Hypothesis: µ = miles
Alternative Hypothesis: µ  miles

61. How to decide whether to reject or accept
How to decide whether to reject or accept ? The entire set of values that the test statistic may assume is divided into two regions. One set, consisting of values that support the and lead to reject , is called the rejection region. The other, consisting of values that support the is called the acceptance region. Tails of a Test
Two-Tailed Test
Left-Tailed Test
Right-Tailed Test
Sign in =
<
>
Rejection Region
In both tail
In the left tail
In the right tail
Rejection Region

62. 2.2.1 a) Testing Hypothesis on the Population Mean,
Null Hypothesis :
Test Statistic :
any population, is known and n is large or
normal population, is known and n is small
any population, is unknown and n is large
normal population, is unknown and n is small

63. Example 2.11

64. Solution

65. Example: Professors’ Salaries
A researcher reports that the average salary of assistant professors is more than $42,000. A sample of 30 assistant professors has a mean salary of $43,260. At α = 0.05, test the claim that assistant professors earn more than $42,000 per year. The standard deviation of the population is $5230.
Step 1: State the hypotheses and identify the claim.
H0: μ = $42,000 and H1: μ > $42,000 (claim)
Step 2: Find the critical value.
Since α = 0.05 and the test is a right-tailed test, the critical value is z = 1.65.

66. Example: Professors’ Salaries
A researcher reports that the average salary of assistant professors is more than $42,000. A sample of 30 assistant professors has a mean salary of $43,260. At α = 0.05, test the claim that assistant professors earn more than $42,000 per year. The standard deviation of the population is $5230.
Step 3: Compute the test value.

67. Example: Professors’ Salaries
Step 4: Make the decision.
Since the test value, 1.32, is less than the critical value, 1.65, and is not in the critical region, the decision is fail to reject the null hypothesis.
Step 5: Summarize the results.
There is not enough evidence to support the claim that assistant professors earn more on average than $42,000 per year.

68. Important Comments
Even though the sample mean of $43,260 is higher than the hypothesized population mean of $42,000, it is not significantly higher. Hence, the difference may be due to chance.
When the null hypothesis is not rejected, there is still a probability of a type II error, i.e., of not rejecting the null hypothesis when it is false.
When the null hypothesis is not rejected, it cannot be accepted as true. There is merely not enough evidence to say that it is false.

69. Definition 2.6: p-value The p-value is the smallest significance level at which the null hypothesis is rejected.

70. Hypothesis Testing
The P-value (or probability value) is the probability of getting a sample statistic (such as the mean) or a more extreme sample statistic in the direction of the alternative hypothesis when the null hypothesis is true.
P-value

71. Hypothesis Testing
In this section, the traditional method for solving hypothesis-testing problems comparesz-values:
critical value
test value
The P-value method for solving hypothesis-testing problems compares areas:
alpha
P-value

72. Solving Hypothesis-Testing Problems
Procedure Table
Solving Hypothesis-Testing Problems
(P-Value Method)
Step 1 State the hypotheses and identify the claim.
Step 2 Compute the test value.
Step 3 Find the P-value.
Step 4 Make the decision.
Step 5 Summarize the results.

73. Example: Cost of College Tuition
A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. Is there evidence to support the claim at a 0.05? Use the P-value method.
Step 1: State the hypotheses and identify the claim.
H0: μ = $5700 and H1: μ > $5700 (claim)

74. Example: Cost of College Tuition
A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. Is there evidence to support the claim at a 0.05? Use the P-value method.
Step 2: Compute the test value.

75. Example: Cost of College Tuition
A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. Is there evidence to support the claim at a 0.05? Use the P-value method.
Step 3: Find the P-value.
Using Table 4, find the area for 0<z < 2.28.
The area is
Subtract from 0.5 to find the area of the tail.
Hence, the P-value is 0.5 – =

76. Example: Cost of College Tuition
Step 4: Make the decision.
Since the P-value is less than 0.05, the decision is to reject the null hypothesis.
Step 5: Summarize the results.
There is enough evidence to support the claim that the tuition and fees at four-year public colleges are greater than $5700.
Note: If α = 0.01, the null hypothesis would not be rejected.

77. Example: Wind Speed
A researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the population is 0.6 mile per hour. At α = 0.05, is there enough evidence to reject the claim? Use the P-value method.
Step 1: State the hypotheses and identify the claim.
H0: μ = 8 (claim) and H1:
Step 2: Compute the test value.

78. Example: Wind Speed
A researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the population is 0.6 mile per hour. At α = 0.05, is there enough evidence to reject the claim? Use the P-value method.
Step 3: Find the P-value.
The area for 0<z <1.89 is
Subtract: 0.5 – =
Since this is a two-tailed test, the area of must be doubled to get the P-value.
The P-value is 2(0.0294) =

79. Example : Wind Speed Step 4: Make the decision.
The decision is to not reject the null hypothesis, since the P-value is greater than 0.05.
Step 5: Summarize the results.
There is not enough evidence to reject the claim that the average wind speed is 8 miles per hour.

80. Guidelines for P-Values With No α
If P-value  0.01, reject the null hypothesis. The difference is highly significant.
If P-value > 0.01 but P-value  0.05, reject the null hypothesis. The difference is significant.
If P-value > 0.05 but P-value  0.10, consider the consequences of type I error before rejecting the null hypothesis.
If P-value > 0.10, do not reject the null hypothesis. The difference is not significant.

81. Significance
The researcher should distinguish between statistical significance and practical significance.
When the null hypothesis is rejected at a specific significance level, it can be concluded that the difference is probably not due to chance and thus is statistically significant. However, the results may not have any practical significance.
It is up to the researcher to use common sense when interpreting the results of a statistical test.

82. b) Hypothesis Test For the Difference between Two Populations Means,
Test statistics:

84. Alternative hypothesis
Rejection Region

85. Example 2.12

86. Solution

88. 2.2.2 a) Testing Hypothesis on the Population Proportion, p
Alternative hypothesis
Rejection Region

89. Example 2.13
When working properly, a machine that is used to make chips for calculators
does not produce more than 4% defective chips. Whenever the machine
produces more than 4% defective chips it needs an adjustment. To check if
the machine is working properly, the quality control department at the
company often takes sample of chips and inspects them to determine if the
chips are good or defective. One such random sample of 200 chips taken
recently from the production line contained 14 defective chips. Test at the 5%
significance level whether or not the machine needs an adjustment.

90. Solution

91. b) Hypothesis Test For the Difference between Two Population Proportion,

92. Alternative hypothesis
Rejection Region

93. Example 2.14:
Reconsider Example 2.5, At the significance level 1%, can we
conclude that the proportion of users of Toothpaste A who will
never switch to another toothpaste is higher than the proportion
of users of Toothpaste B who will never switch to another
toothpaste?

94. Solution

95. Additional Topics Regarding Hypothesis Testing
There is a relationship between confidence intervals and hypothesis testing.
When the null hypothesis is rejected in a hypothesis-testing situation, the confidence interval for the mean using the same level of significance will not contain the hypothesized mean.
Likewise, when the null hypothesis is not rejected, the confidence interval computed using the same level of significance will contain the hypothesized mean.

96. Example: Sugar Production
Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. Is there enough evidence to conclude that the bags do not contain 5 pounds as stated at α = 0.05? Also, find the 95% confidence interval of the true mean.
Step 1: State the hypotheses and identify the claim.
H0: μ = 5 and H1: μ  5 (claim)
Step 2: Find the critical value.
The critical values are z = ±1.96.

97. Example: Sugar Production
Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. Is there enough evidence to conclude that the bags do not contain 5 pounds as stated at α = 0.05? Also, find the 95% confidence interval of the true mean.
Step 3: Compute the test value.

98. Example: Sugar Production
Step 4: Make the decision.
Reject the null hypothesis.
Step 5: Summarize the results.
There is enough evidence to support the claim that the bags do not weigh 5 pounds.

99. Example: Sugar Production
The 95% confidence interval for the mean is
Notice that the 95% confidence interval of m does not contain the hypothesized value μ = 5.
Hence, there is agreement between the hypothesis test and the confidence interval.

100. Power of a Statistical Test
The power of a test measures the sensitivity of the test to detect a real difference in parameters if one actually exists. The higher the power, the more sensitive the test. The power is 1 – β.