1 definitions in statistics, a hypothesis is a claim or statement about a property of a population....

1

Definitions

In statistics, a hypothesis is a claim or statement about a property of a population.

A hypothesis test is a standard procedure for testing a claim about a property of a population.

2

Main Objectives

We will study hypothesis testing for

1. population proportion p

2. population mean 3. population standard deviation

3

ExampleClaim: the XSORT method of gender

selection increases the likelihood of having a baby girl.

This is a claim about proportion (of girls)

To test this claim 14 couples (volunteers) were subject to XSORT treatment.

If 6 or 7 or 8 have girls, the method probably does not increase the likelihood of a girl.

If 13 or 14 couples have girls, the method is probably increases the likelihood of a girl.

4

Rare Event Rule for Inferential Statistics

If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct.

5

Components of aFormal

Hypothesis Test

6

Null Hypothesis: H0

• The null hypothesis (denoted by H0) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value.

• We test the null hypothesis directly.

• Either reject H0 or fail to reject H0

(in other words, accept H0 ).

7

Alternative Hypothesis: H1

• The alternative hypothesis (denoted by H1) is the statement that the parameter has a value that somehow differs from the null hypothesis.

• The symbolic form of the alternative hypothesis must use one of these symbols: , <, >. (not equal, less than, greater than)

8

Example 1

Claim: the XSORT method of gender selection increases the likelihood of having a baby girl. We express this claim in symbolic form: p>0.5 (here p denotes the proportion of baby girls)

Null hypothesis must say “equal to”, so H0 : p=0.5Alternative hypothesis must express difference: H1 : p>0.5Original claim is now the alternative hypothesis

9

Example 1 (continued)

We always test the null hypothesis.

If we reject the null hypothesis, then the original clam is accepted.

Final conclusion would be: XSORT method increases the likelihood of having a baby girl.

If we fail to reject the null hypothesis, then the original clam is rejected.

Final conclusion would be: XSORT method does not increase the likelihood of having a baby girl.

10

Example 2

Claim: for couples using the XSORT method the likelihood of having a baby girl is 50% Express this claim in symbolic form: p=0.5 (again p denotes the proportion of baby girls)

Null hypothesis must say “equal to”, so H0 : p=0.5Alternative hypothesis must express difference: H1 : p0.5Original claim is now the null hypothesis

11

Example 2 (continued)

If we reject the null hypothesis, then the original clam is rejected.

Final conclusion would be: for couples using the XSORT, the likelihood of having a baby girl is not 0.5

If we fail to reject the null hypothesis, then the original clam is accepted.

Final conclusion would be: for couples using the XSORT the likelihood of having a baby girl is indeed equal to 0.5

12

Example 3

Claim: for couples using the XSORT method the likelihood of having a baby girl is at least 0.5 Express this claim in symbolic form: p≥0.5 (again p denotes the proportion of baby girls)

Null hypothesis must say “equal to”, so H0 : p=0.5 (this agrees with the claim!)

Alternative hypothesis must express difference: H1 : p<0.5Original claim is now the null hypothesis

13

Example 3 (continued)

If we reject the null hypothesis, then the original clam is rejected.

Final conclusion would be: for couples using the XSORT, the likelihood of having a baby girl is less 0.5

If we fail to reject the null hypothesis, then the original clam is accepted.

Final conclusion would be: for couples using the XSORT the likelihood of having a baby girl is indeed at least 0.5

14

General rules:

• If the null hypothesis is rejected, the alternative hypothesis is accepted.

• If the null hypothesis is accepted, the alternative hypothesis is rejected.

• Acceptance or rejection of the null hypothesis is an initial conclusion.

• Always state the final conclusion expressed in terms of the original claim, not in terms of the null hypothesis or the alternative hypothesis.

15

Type I Error

• A Type I error is the mistake of rejecting the null hypothesis when it is actually true.

• The symbol (alpha) is used to represent the probability of a type I error.

16

Type II Error

• A Type II error is the mistake of accepting the null hypothesis when it is actually false.

• The symbol (beta) is used to represent the probability of a type II error.

17

Type I and Type II Errors

18

Example

Claim: a new medicine has a greater success rate, p>p0, than the old (existing) one.

Null hypothesis: H0 : p=p0

Alternative hypothesis: H1 : p>p0 (agrees with the original claim)

19

Example (continued)

Type I error: the null hypothesis is true, but we reject it => we accept the claim, hence we adopt the new (inefficient, potentially harmful) medicine.

This is a critical error, should be avoided!

Type II error: the alternative hypothesis is true, but we reject it => we reject the claim, hence we decline the new medicine and continue using the old one (no harm…).

20

Significance Level

The probability of the type I error (denoted by ) is also called the significance level of the test.

It characterizes the chances that the test fails (i.e., type I error occurs)

It must be a small number. Typical values used in practice: = 0.1, 0.05, or 0.01 (in percents, 10%, 5%, or 1%).

21

The test statistic is a value used in making a decision about the null hypothesis.

The test statistic is computed by a specific formula depending on the type of the test.

Testing hypothesis Step 1: compute Test Statistic

22

Section 8-3

Testing a Claim About a Proportion

23

Notation

p = population proportion (must be

specified in the null hypothesis)

q = 1 – p

n = number of trials

p = (sample proportion)xn

24

1) The sample observations are a simple random sample.

2) The conditions for a binomial distribution are satisfied.

3) The conditions np 5 and nq 5 are both satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with µ = np and = npq . Note: p is the assumed proportion not the sample proportion.

Requirements for Testing Claims About a Population Proportion p

25

p – p

pqn

z =

Test Statistic for Testing a Claim About a Proportion

Note: p is the value specified in the null hypothesis; q = 1-p

26

Example 1 again:

Claim: the XSORT method of gender selection increases the likelihood of having a baby girl.

Null hypothesis: H0 : p=0.5Alternative hypothesis: H1 : p>0.5

Suppose 14 couples treated by XSORT gave birth to 13 girls and 1 boy.

Test the claim at a 5% significance level

27

Compute the test statistic:

p̂ 13 14 0.929

z p̂ p

pqn

0.929 0.5

0.5 0.5 14

3.21

28

Draw the diagram (the normal curve)

On the diagram, mark a region of extreme values that agree with the alternative hypothesis:

Sample proportion of: orTest Statistic z = 3.21

p̂ 0.929

29

Critical Region

The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis.

For example, see the red-shaded region in the previous figure.

30

Critical Value

A critical value is a value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis.

See the previous figure where the critical value is z = 1.645. It corresponds to a significance level of = 0.05.

31

Significance Level

The significance level (denoted by ) is the probability that the test statistic will fall in the critical region (when the null hypothesis is actually true).

32

Conclusion of the test

Since the test statistic (z=3.21) falls in the critical region (z>1.645), we reject the null hypothesis.

Final conclusion: the original claim is accepted, the XSORT method of gender selection indeed increases the likelihood of having a baby girl.

33

Types of Hypothesis Tests:Two-tailed, Left-tailed, Right-tailed

The tails in a distribution are the extreme regions where values of the test statistic agree with the alternative hypothesis

34

Right-tailed Test

H0: p=0.5

H1: p>0.5Points Right

is in the right tail

35

Critical value for a right-tailed test

A right-tailed test requires one (positive) critical value:

z

36

Left-tailed Test

H0: p=0.5

H1: p<0.5Points Left

is in the left tail

37

Critical value for a left-tailed test

A left-tailed test requires one (negative) critical value:

─z

38

Two-tailed Test

H0: p=0.5

H1: p0.5

is divided equally between the two tails of the critical

region

Means less than or greater than

39

Critical values for a two-tailed test

A two-tailed test requires two critical values:

z/2 and ─z/2

40

P-Value

The P-value (or p-value or probability value) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true.

41

Example 1 (continued)

P-value is the area to the right of the test statistic z = 3.21.

We refer to Table A-2 (or use calculator) to find that the area to the right of z = 3.21 is 0.0007.

P-value = 0.0007

42

P-Value method:

If the P is low, the null must go.If the P is high, the null will fly.

If P-value , reject H0.

If P-value > , fail to reject H0.

43

Example 1 (continued)

P-value = 0.0007

It is smaller than = 0.05.

Hence the null hypothesis must be rejected

44

P-Value

Critical region in the left tail:

Critical region in the right tail:

Critical region in two tails:

P-value = area to the left of the test statistic

P-value = area to the right of the test statistic

P-value = twice the area in the tail beyond the test statistic (see the following diagram)

45

Procedure for Finding P-ValuesFigure 8-5

46

Caution

Don’t confuse a P-value with a proportion p.Know this distinction:

P-value = probability of getting a test statistic at least as extreme as the one representing sample data

p = population proportion

47

If the test statistic falls within the critical region, reject H0.

If the test statistic does not fall within the critical region, fail to reject H0 (i.e., accept H0).

Traditional method:

48

P-Value method:

If the P is low, the null must go.If the P is high, the null will fly.

If P-value is small ( ), reject H0.

If P-value is not small (> ), accept H0.

49

• Press STAT and select TESTS

• Scroll down to 1-PropZTest press ENTER

• Type in p0: (claimed proportion, from H0)

• x: (number of successes)

• n: (number of trials)

• choose H1: p ≠p0 <p0 >p0

(two tails) (left tail) (right tail)

• Press on Calculate

• Read the test statistic z=…

• and the P-value p=…

Testing hypothesis by TI-83/84

50

Do we prove a claim?

• A statistical test cannot prove a hypothesis or a claim.

• Our conclusion can be only stated like this: the available evidence is not strong enough to warrant rejection of a hypothesis or a claim (such as not enough evidence to convict a suspect).

51

Section 8-4

Testing a Claim About a Mean: Known

52

Notation

n = sample size

= sample mean

= claimed population mean (from H0)

= known value of the population standard deviation

x

53

Requirements for Testing Claims About a Population Mean (with Known)

1) The value of the population standard deviation is known.

2) Either or both of these conditions is satisfied: The population is normally distributed or n > 30.

54

Test Statistic for Testing a Claim About a Mean (with Known)

n

x – µz =

55

Example:People have died in boat accidents because an obsolete estimate of the mean weight of men (166.3 lb) was used.

A random sample of n = 40 men yielded the mean = 172.55 lb. Research from other sources suggests that the population of weights of men has a standard deviation given by = 26 lb.

Test the claim that men have a mean weight greater than 166.3 lb.

x

56

Example:Requirements are satisfied: is known (26 lb), sample size is 40 (n > 30)

We can express claim as > 166.3 lb

It does not contain equality, so it is the alternative hypothesis.

H0: = 166.3 lb null hypothesisH1: > 166.3 lb alternative hypothesis

(and original claim)

57

Example:

Let us set significance level to = 0.05

Next we calculate z

z x

x

n

172.55 166.3

26

40

1.52

It is a right-tailed test, so P-value is the area is to the right of z = 1.52;

58

Example:Table A-2: area to the left of z = 1.52 is 0.9357, so the area to the right is1 – 0.9357 = 0.0643.The P-value is 0.0643

The P-value of 0.0643 is greater than the significance level of = 0.05, we fail to reject the null hypothesis.

x 172.55

= 166.3

orz = 0 or

z = 1.52

P-value = 0.0643

59

Example:

The traditional method:

Use critical value z = 1.645 instead of finding the P-value. Since z = 1.52 does not fall in the critical region, again fail to reject the null hypothesis.

60

• Press STAT and select TESTS

• Scroll down to Z-Test press ENTER

• Choose Data or Stats. For Stats:

• Type in 0: (claimed mean, from H0)

• : (known st. deviation)

• x: (sample mean)

• n: (sample size)

choose H1: ≠0 <0 >0

(two tails) (left tail) (right tail)

Testing hypothesis by TI-83/84

61

• (continued)

• Press on Calculate

• Read the test statistic z=…

• and the P-value p=…

62

Section 8-5

Testing a Claim About a Mean: Not Known

63

Notation

n = sample size

= sample mean

= claimed population mean (from H0)

s = sample standard deviation

x

64

Requirements for Testing Claims About a Population

Mean (with Not Known)

1) The value of the population standard deviation is not known.

2) Either or both of these conditions is satisfied: The population is normally distributed or n > 30.

65

Test Statistic for Testing a Claim About a Mean (with Not Known)

P-values and Critical ValuesFound in Table A-3 or by calculator

Degrees of freedom (df) = n – 1

x – µt = s

n

66

Example:People have died in boat accidents because an obsolete estimate of the mean weight of men (166.3 lb) was used.

A random sample of n = 40 men yielded the mean = 172.55 lb and standard deviation s = 26.33 lb. Do not assume that the population standard deviation is known.

Test the claim that men have a mean weight greater than 166.3 lb.

x

67

Example:Requirements are satisfied: is not known, sample size is 40 (n > 30)

We can express claim as > 166.3 lb

It does not contain equality, so it is the alternative hypothesis.

H0: = 166.3 lb null hypothesisH1: > 166.3 lb alternative hypothesis

(and original claim)

68

Example:

Let us set significance level to = 0.05

Next we calculate t

t x

x

s

n

172.55 166.3

26.33

40

1.501

df = n – 1 = 39area of 0.05, one-tail yields critical value t = 1.685;

69

Example:

t = 1.501 does not fall in the critical region bounded by t = 1.685, we fail to reject the null hypothesis.

= 166.3

orz = 0

x 172.55or

t = 1.52

Critical value t =

1.685

70

Example:

Final conclusion:

Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support a conclusion that the population mean is greater than 166.3 lb.

71

The critical value in the preceding example was t = 1.782, but if the normal distribution were being used, the critical value would have been z = 1.645.

The Student t critical value is larger (farther to the right), showing that with the Student t distribution, the sample evidence must be more extreme before we can consider it to be significant.

Normal Distribution Versus Student t Distribution

72

P-Value Method

Use software or a TI-83/84 Plus calculator.

If technology is not available, use Table A-3 to identify a range of P-values

(this will be explained in Section 8.6)

73

• Press STAT and select TESTS

• Scroll down to T-Test press ENTER

• Choose Data or Stats. For Stats:

• Type in 0: (claimed mean, from H0)

• x: (sample mean)

• sx: (sample st. deviation)

• n: (sample size)

• choose H1: ≠0 <0 >0

(two tails) (left tail) (right tail)

Testing hypothesis by TI-83/84

74

• (continued)

• Press on Calculate

• Read the test statistic t=…

• and the P-value p=…

75

Section 8-6

Testing a Claim About a Standard Deviation or

Variance

76

Notation

n = sample size

s = sample standard deviation

s2 = sample variance

= claimed value of the population

standard deviation (from H0 )

2 = claimed value of the population

variance (from H0 )

77

Requirements for Testing Claims About or 2

1. The sample is a simple random sample.

2. The population has a normal distribution. (This is a much stricter requirement than the requirement of a normal distribution when testing claims about means.)

78

Chi-Square Distribution

Test Statistic

2

n 1 s2

2

79

Critical Values for Chi-Square Distribution

• Use Table A-4.

• The degrees of freedom df = n –1.

80

Table A-4Table A-4 is based on cumulative areas from the right.

Critical values are found in Table A-4 by first locating the row corresponding to the appropriate number of degrees of freedom (where df = n –1).

Next, the significance level is used to determine the correct column.

The following examples are based on a significance level of = 0.05.

81

Critical values Right-tailed test: needs one critical value

Because the area to the right of the critical value is 0.05, locate 0.05 at the top of Table A-4.

Area Area 1-

critical value

82

Critical valuesLeft-tailed test: needs one critical value

With a left-tailed area of 0.05, the area to the right of the critical value is 0.95, so locate 0.95 at the top of Table A-4.

Area Area 1-

1-

critical value

83

Critical valuesTwo-tailed test: needs two critical values

Critical values are two different positive numbers, both taken from Table A-4

Divide a significance level of 0.05 between the left and right tails, so the areas to the right of the two critical values are 0.975 and 0.025, respectively.

Locate 0.975 and 0.025 at top of Table A-4

84

Area Area 1-

right critical value

Area

left critical value

Critical values for a two-tailed test

85

Finding a range for P-value

• A useful interpretation of the P-value: it is observed level of significance.

• Compare your test statistic 2 with critical values shown in Table A-4 on the line with df=n-1 degrees of freedom.

• Find the two critical values that enclose your test statistic. Determine the significance levels 1 and 2 for those two critical values.

• Your P-value is between 1 and 2

(see examples below)

86

Examples: Right-tailed test: If the test statistic 2 is between critical values corresponding to the areas 1 and 2 , then your P-value is between 1 and 2 .

Left-tailed test: If the test statistic 2 is between critical values corresponding to the areas 1-1 and 1-2 , then your P-value is between 1 and 2 .

87

Examples: Two-tailed test: If the test statistic 2 is between critical values corresponding to the areas 1 and 2 , then your P-value is between 21 and 22 .

Two-tailed test: If the test statistic 2 is between critical values corresponding to the areas 1-1 and 1-2 , then your P-value is between 21 and 22 .

(Note: for two-tailed tests, multiply the areas by two)

88

Finding the exact P-value by TI-83/84

• Use the test statistic 2 and the calculator function 2

cdf to compute the area of the tail:

2 cdf(teststat,999,df) gives the area of the right tail (to the right from the test statistic)

2 cdf(-999,teststat,df) gives the area of the left tail (to the left from the test statistic)

Multiply the area of the tail by 2 if you have a two-tailed test