Chapter 2 Hypothesis Testing

Chapter 2 Statistical Hypothesis Testing

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STATISTICAL TESTING OF HYPOTHESIS

Concepts and Definitions:

In reality, the problem confronting the scientist or researcher is not so much the estimation of a population parameter, but rather the formation of a data-based decision procedure that can produce a conclusion about some scientific system. For example, the BFAD must decide whether a new flu vaccine is “effective” or “ineffective”, or a sociologist might wish to collect between appropriate data to enable him or her to decide whether a person’s blood type and skin color are independent variables. In each of these cases the BFAD or the sociologist postulates or conjectures something about a system. In addition, each must involve the use of experimental data and decision making that is based on the data. Formally, in each case, the postulate can be put in the form of statistical hypothesis. Procedures that lead to the acceptance or rejection of statistical hypotheses such as these comprise a major role of statistical inference. In this section, we are going to discuss the basic principles of statistical hypothesis testing.

What is a Hypothesis?

A hypothesis is an assertion or a statement about a population. Data are gathered to check the validity of this claim. In practice, it is assumed that the population is so large that it is not feasible to collect all its elements and verify the hypothesis. In this light, we only use the available data (large enough) to check the reasonableness of the statement.

What is Hypothesis Testing?

Statistical hypothesis testing is a procedure of dichotomizing the conflicting theories, and test these statements based from the sample evidence and probability theory. At the end of this procedure, we can “reject” or “do not reject” the assumed true hypothesis.

Five-Step Procedure for Testing a Hypothesis

There is a five-step procedure that systematizes hypothesis testing; when we get to step 5, we are ready to reject or not reject the hypothesis. But keep in mind that this procedure used by statistician does not provide proof that something is true, in the manner that mathematician “proves” a theorem. It does provide a “proof beyond reasonable doubt,” in the manner of the court system. The reader must be accustomed to understand that the acceptance of a hypothesis merely implies that the data do not give sufficient evidence to refute it. On the other way, rejection implies that the sample evidence refutes it. Put another way, rejection means that there is a small probability of obtaining the sample information observed when, in fact, the hypothesis is true.

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__________________________________________________________________Step 1. State the Null Hypothesis and the Alternative Hypothesis.

This procedure always starts with giving the two mutually exhaustive hypotheses: the null hypothesis, denoted by , and the alternative hypothesis,

denoted by . The null hypothesis is always assumed to be true before performing the test. Sometimes, it is referred as the status quo. In practice, the null hypothesis is expressed as a statement concerning the value of a population parameter (say the mean). On the other hand, the alternative hypothesis describes what you will conclude if you reject the null hypothesis. It is often called the research hypothesis since it is the alternative that researchers want to get.

To illustrate these concepts, suppose we want to test whether a coin is “fair” or “biased”. Based from this problem, the appropriate null hypothesis will be the probability of getting a head, say p, must be equal to ½. That is,

. On the other hand, the appropriate alternative hypothesis will be . The alternative seems obvious! The coin will be biased if the chance of getting a head or tail is not equal to 0.5. Remember that in most cases, the null hypothesis is expressed as , where is the unknown population parameter and is the assumed value to be tested.

Step 2. Select a Level of Significance.

After establishing the null and alternative hypotheses, the next step is to choose the level of significance, denoted by . To understand the notion of level of significance, a researcher may commit two possible errors in testing hypothesis. If the researcher rejects the null hypothesis, given that it is true, then he or she committed a Type I error. On the other hand, if the researcher does not reject the null hypothesis, given that it is false, then he or she committed a Type II error. The following table below summarizes the decisions the researcher could make and the possible consequences.

Researcher

Null hypothesis Do not reject H_0 Rejects H_0

H_0 is true Correct decision Type I error

H_0 is false Type II error Correct decision

We denote , while . In theory,

both of these error probabilities must be small, in order to say that the test is valid. But simulation studies show that we cannot minimize these error probabilities simultaneously. In fact, if we decrease the , then increases. The reverse process is also true. In practice, we usually select a small value for , and find a test with maximum . In statistical theory, is called the power of a test. In biological studies, the usual values of are 0.05 and 0.01. Sometimes, a test that rejects the null hypothesis at 0.05 level of significance is called significant,

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__________________________________________________________________while a test that rejects the null hypothesis at 0.01 level of significance is called highly significant.Step 3. Select the Appropriate Test Statistic

A test statistic is a value, determined from sample information, used to determine whether to reject the null hypothesis. We will deal with these in details later.

Now consider the tossing of a coin problem posted earlier. Since we want to test the “fairness” of the coin, the appropriate test statistic is the number of heads occurred in n tosses, say n = 100. We can also use the ratio between the number of heads occurred and the total number of tosses.

Step 4. Formulate the Decision Rule.

A decision rule is a statement of the specific conditions under which the null hypothesis is rejected and the conditions under which it is not rejected. The region or area of rejection defined the location of all those values that are so large or so small that the probability of their occurrence under true null hypothesis is rather remote.

Going back to our problem regarding the “fairness” of a coin, note that if the null hypothesis is true (i.e., p = 0.5), then the number of heads in 100 tosses should not be too different from 50. So if the decision rule is to reject if and only if or , then we tend to reject the null hypothesis very often. This is the same as saying that a defendant is guilty if one witness appears in the court! On the other extreme, if we reject the null hypothesis if and only if or , then we can think of this as having a roomful of witnesses before we say that the defendant is guilty. In practice, the decision rule will be based on the value of and the probability distribution of the test statistic.

Step 5. Make a DecisionThe fifth and final step is computing the value of the test statistic, check

whether its value is inside the rejection region, and making a decision to reject or not reject the null hypothesis.

P-values in Hypothesis Test

A commonly used method in testing hypothesis is to report the p-value of the data. A p-value does not require imposing a pre-selected level of significance. (Sometimes, selecting the appropriate is troublesome to the researcher.) The p-value is the probability that the test statistic will take on a value that is at least as extreme as the observed value of the statistic when the null hypothesis is assumed to be true. Thus, a p-value conveys much information about the weight of evidence against and so a decision maker can draw a conclusion at any specified level of significance. Formally, we define the p-value as the smallest level of significance that would lead to the rejection of the null hypothesis given the observed data. With this definition, if the observed p-value is equal to 0.40, then we will reject the null hypothesis at this level. But using this value is too large! We can’t afford to have this very large probability of committing type I error. Thus, we tend to do not reject the null hypothesis. On the other hand, if p-value = 0.0001, then we can reject the null hypothesis at this very small level of significance. In this case, we reject the null hypothesis. Thus, given a specified value , we reject if and only if . Otherwise, we do not reject.

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__________________________________________________________________This method of testing hypothesis is very popular nowadays since almost

all statistical packages report the p-value for specific test.

One-Tailed and Two-Tailed Tests

In theory, it is assumed that the test statistic has no fixed value. That is, it is assumed to be random. The probability distribution of a test statistic is often called as the sampling distribution. Now in hypothesis testing, we partition the set of all values of the test statistic into rejection and acceptance region. If the value of the test statistic falls in the rejection region, then we reject the null hypothesis. Otherwise, we do not reject it. If the region rejections are located at the tails of the distribution of the test statistic, then we have a two-tailed test.

Otherwise, we have a one-tailed test. The figure below depicts two one-tailed tests using z statistic.

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(Left) one-tailed test

(Right) one-tailed test


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Tests Concerning a Single Mean Large Sample ( Z –test )

Suppose that a random sample is drawn from a normal population with mean and standard deviation , and suppose we want to test

against an alternative hypothesis . If is known, then the test statistic for this testing problem is given by

This test statistic has a standard normal distribution, and often called the z-test. For a specified level of significance , the rejection region for specific alternative hypothesis is given below.

Alternative hypothesis Reject if and only if

or

In the table, the critical value is chosen such that .

Example 1: Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient’s body, but the battery pack needs to be recharged about every four hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation hours. Is there evidence to support the claim that the mean battery life exceeds 4 hours? Use .

Solution: We will solve this testing problem in 5 steps. 1. (the mean life time of the battery packs is 4 hours)

(the mean life time of the battery packs exceeds 4 hours)

2. In the problem, it is indicated to use 0.05 level of significance.

3. Since we are testing the mean lifetime and the population standard deviation is known ( / n = 0.2 / 50 = 0.02828 ), then we use megastat. In the normal distribution box, enter 4 for the mean,

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__________________________________________________________________0.02828 for the standard deviation, and 4.05 for the x. Click preview for the result. The result is z = 1.77.

4. At 0.05 level of significance, we reject if and only if

. (Using normal distribution by megastat or NORMSINV function ).

NOTE: Since it is a right tailed test, use positive result. Thus, Z > z 0.05 = 1.645

5. Based from the rejection region (and critical value 1.645), we see that the z value is inside the rejection region (or 1.77 is greater than 1.645). Thus, we reject the null hypothesis and conclude that the battery life exceeds 4 hours.

Example 2 A random sample of 100 deaths in the Philippines last year showed an average life span of 69.3 years. Assuming a population standard deviation of 7.8 years, does this seem to indicate that the life span today is lesser than 70 years? Use a 0.01 level of significance.

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1.645Rejection region for the example

1.77


__________________________________________________________________Solution: We will solve this testing problem in 5 steps.

1. H0: = 70 years.H1: < 70 years.

2. Use a = 0.01.

3. Since we are testing the mean life span and the population standard deviation is known ( / n = 7.8 / 100 = 0.78. Using normal distribution under megastat, z = – 0.90.

4. At 0.01 level of significance, we reject if and only if Z < z 0.01= -2.33 (Using normal distribution by megastat or NORMSINV function)

Note: Since it is a left-tailed test, use negative result. Thus, z = -2.33.

5. Based from the rejection region (and critical value –2.33), we see that the z value is outside the rejection region (or –0.90 is greater than –2.33). Thus, we do not reject the null hypothesis and conclude that the life expectancy of Filipinos is 70 years.

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-2.33

-0.90


__________________________________________________________________Other Uses for the Normal Z Test

The normal Z test can also be used on the following cases:1. the population standard deviation is unknown and the sample size is

large enough. (Simulations show that the approximation is valid provided that the sample size is at least 30.) In this case, we replace by s, where s

is defined by .

2. the population is not known to be normal and the sample size is large enough. In this case, we apply the central limit theorem. Also, if the population standard deviation is not known, use s.

Small Sample ( t –test )

Another important continuous distribution in statistics: the Student t distribution or simply the t distribution. The t curve and the standard normal curve are shown below.

Z distribution T distribution

Note particularly that the t distribution is flatter, more spread out, than the standard normal distribution. This is because t distributions have larger standard deviations than the standard normal.

The following characteristics of the t distribution are based on the assumption that the population of interest is normal, or nearly normal.

1. It is, like the z distribution, bell-shaped and symmetrical about zero (0).

2. There is not one t distribution, but rather a “family” of t distribution. All t distributions have mean 0, but their standard deviations differ according to the sample size n. But as the sample size increases, the t distribution approaches the z distribution. That is, for large n, t and z are almost identical.

Suppose that we observe a random sample from a normal population with mean and standard deviation . In this case, we assume that the sample size is small (usually smaller than 30) and the population standard deviation is not known (In practice this is usually the case). Suppose we want to test versus some alternative hypothesis. For this testing problem, we use the t statistic defined by

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__________________________________________________________________where S is the sample standard deviation. The following provides the rejection regions for this testing problem.

Alternative Hypothesis Rejection Region

or

where is the critical value of a t distribution with degrees of freedom

such that .

Example 3 A recent survey stated that cell phone owners received an average of 50 texts daily. To test the claim, a researcher surveyed 25 cell phone owners and found out that the average number of received text was 46. The standard deviation of the sample was 7. At 0.03 level of significance, is there enough evidence to reject the survey’s claim?

Solution: 1. H0: = 50H1: 50

2. Use a = 0.03.

3. Compute for the t – statistic. Using Excel, enter the label ( text ), sample mean ( 46 ), sample s.d. ( 7 ), and number of samples ( 25 ).Click megastat menu, then select Hypothesis tests and choose Mean vs. Hypothesized Value. At the dialog box, select summary input. In the input range, highlight the range values from A1 to A4. Enter 50, the population mean, for the hypothesized mean, not equal for the alternative, and 97% for the confidence interval. Note that confidence interval is equal to ( 1 – a ) 100%. Choose, of course, t-test. For clarity, look at the figure below.

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After pressing OK, megastat will provide an output sheet where all the necessary computations and statistics will be displayed.

The result t – value = - 2.86.

4. Compute for the critical region. At a = 0.03, we reject H0 if or . Using t – distribution, select

calculate t given probability. Enter 0.015 for the probability for it is two-tailed test ( a/2 = 0.03/2 = 0.015 ). Enter 24 for the degrees of freedom ( n – 1 = 25 – 1 = 24 ). Click preview.

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Note: Since it is two-tailed, t = 2.31.

5. Based from the rejection region (and critical value 2.31), we see that the t value is inside the rejection region (or –2.86 < - 2.31 ). Thus, we reject the null hypothesis and conclude that cell phone owners received text is not equal to 50 texts daily.

Example 4 A company claims that the mean weight per banana it ships is 150 grams. A quality control supervisor, inspect from a random sample of 11 and weigh each banana. The results are reported below in grams.

Is there a sufficient evidence to reject the company’s claim? Use a = 0.05.

Solution: 1. H0 = 150 grams.H1 150

2. Use a = 0.05.

3. Compute for the t – statistic. Using Excel, enter the data. Click megastat menu, then select Hypothesis tests and choose Mean vs. Hypothesized Value. At the dialog box, select data input. In the input range, highlight the range values from A1 to A11. Enter 150, the population mean, for the hypothesized mean, not equal for the alternative, and 95% for the confidence interval. Note that confidence interval is equal to ( 1 – a ) 100%. Choose, of course, t-test. For clarity, look at the figure below.

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Rejection region- 2.86

-2.31 2.31

152 149 157 155 152 148 147 149 150 152 156


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The result t – value = 1.54.

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4. The critical region using t – distribution, t = 2.23.

5. Do not reject. Therefore, the mean weight of banana was 150 grams.

Two Samples: Tests on Two Means

Normal Test for Two Independent Samples

Suppose now that we obtain two independent normal samples. That is, let be normally distributed with mean and standard deviation, and

be normally distributed with mean and standard deviation . In this

case, we want to test versus some alternative hypothesis . If the population standard deviations are both known, then we use the Z statistic defined by

The rejection regions, together with their respective alternative hypothesis, are summarized below.


or

Note that the above test can also be used when the population standard deviations are not known provided that both n and m are large. Just replace by

, and by .

Example 5 A soft-drink manufacturer claims that its 12 – ounce bottles contains on average 30 calories. A supervisor took a random sample of 40 bottles of this soft-drink, which were checked for calories, contained a mean of 32.4 calories and a standard deviation of 2 calories. Another supervisor took a random sample of 50 bottles and found out that the mean calories contained were 29.5 and standard deviation of 3.4 calories. Test at 1% significance level if the sample results indicates that the bottled soft-drink have equal calories content.

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-2.23 2.23

1.54


__________________________________________________________________Solution: 1. ( samples have the same calorie content )

H1: 1 2

2. Use a = 0.01 level of significance.

3. Compute for the z– statistic. Using excel, enter the data intotwo columns. In this case consider the following data for column A: test 1, 32.4, 2, 40; and for column B: test 2, 29.5, 3.4, 50. Click megastat menu and select Hypothesis Tests. In the Hypothesis Tests option, select Compare Two Independent Groups. At the dialog box, select summary input. Highlight A1 to A4 for Group 1 and B1 to B4 for Group 2. Select z – test option and change the confidence interval for 99%. Look at the figure on the next page


The result is z = 5.04.

4. The critical region at a = 0.01 ( two tailed ), z = 2.58.

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5. Reject H0 and conclude that the samples don’t have the same calorie content.

Two-Sample t -Test

In this case, we consider two independent random samples and

, where is normally distributed with mean and standard

deviation , while is normally distributed with mean and standard deviation

Also, we assume that both the sample sizes n and m are small. Suppose we

consider testing versus some alternative hypothesis. If we assume that the two population standard deviations are equal, then the test statistic for this testing problem is given by

where , known as the pooled standard deviation, is given by

The t-statistic has a t distribution with degrees of freedom. The following table summarizes the rejection regions and the corresponding alternative hypotheses.


or

Example 6 A manufacturer claims that the average tensile strength of thread X exceeds the average tensile strength of thread Y by at least 10 kilograms. To test his claim, 17 pieces of each type of thread are tested under similar conditions. Type X thread had an average tensile strength of 85.7 kilograms with standard deviation of 5.67 kilograms, while thread Y had an average tensile strength of 75.3 kilograms with standard deviation of 4.46 kilograms. Test the manufacturer’s claim using a 0.01 level of significance.

Solution: 1. H0: 1 2 = 10.H1: 1 2 > 10.

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-2.58 2.58

5.04


__________________________________________________________________2. Use a = 0.01.

3. Compute for the t– statistic. Using excel, enter the data intotwo columns. In this case consider the following data for column A: thread X, 85.7, 5.67, 17; and for column B: thread Y, 75.3, 4.46, 17. Click megastat menu and select Hypothesis Tests. In the Hypothesis Tests option, select Compare Two Independent Groups. At the dialog box, select summary input. Highlight A1 to A4 for Group 1 and B1 to B4 for Group 2. Enter 10 at the hypothesized difference. Select greater than option in the Alternative. Select t – test ( pooled variance ) option and change the confidence interval for 99%.

Press OK for result.

The result, t = 0.23.

4. The critical region at a = 0.01 with 32 degrees of freedom is t > 2.45.

32

0.23


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5. Do not reject. We are unable to conclude that the tensile strength of thread X exceeds that of thread Y by more than 10 kilograms.

Test on a Single Proportion

Tests of hypotheses concerning proportions are required in many areas. All manufacturing firms are concerned about the proportion of defective items when a shipment is made. The gambler depends on knowledge of the proportion of outcomes that he considers favorable. A politician is certainly interested in knowing what fractions of the voters will favor him in the next election.

For small sample size, the binomial test is the appropriate test. However, several studies showed that the normal distribution provides a very accurate approximation to the binomial distribution when n is large and p is close to ½.. Thus, the z-test which is the transformation of binomial random variable with mean = np and standard deviation = npq. The limiting form of the distribution

Z =

Example 7 A politician claims that he will garner 90% votes from his bailiwick province. Would you agree to his claim if on a given day a researcher asked 1000 qualified voters and turn out that only 872 is in favor of the politician? Use a = 0.05

Solution: 1) H0: p = 0.9H1: p 0.9

2) Let a = 0.05

3) Computation:

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2.45

x – np npq


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z = -2.95; p-value = 0.00324) Critical Region

Z 0.05 / 2 = 1.96

5) Decision: Reject H0.

Conclusion: We have reason to doubt the politician’s claim.

Test on Two Proportions

A person may decide to give up drinking only if he is convinced that the proportion of drinkers with liver cancers the proportion of nondrinkers with lung cancers. Situations arise where we wish to test the hypothesis that two proportions are equal. In general, we wish to test the null hypothesis that two proportions, or binomial parameters, are equal.

Example 8 In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a thermal power plant, it is found that 73 out of 130 urban residents favor the construction while only 60 of 150 suburban residents are in favor. Is there significant difference between the proportion of urban and suburban residents who favor construction of the thermal plant? Use a 0.01 level of significance.

Solution: 1) H0: p1 = p2

H1: p1 p2

2) Let a = 0.01.

3) Computation:

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z = 2.70; p-value = 0.0069

4) Critical Region

Z 0.01 / 2 = 2.58


Conclusion: The proportion of urban residents favoring the construction is not equal to the proportion of suburban residents

One – and – Two – Sample Tests Concerning Variances

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__________________________________________________________________Engineers and scientists are constantly confronted with studies in which

they are required to demonstrate that measurements involving products or processes fall inside specifications set by consumers. The specifications are often met if the process variance is sufficiently small. It is focused on comparative experiments between processes where inherent variability must formally be compared. A test comparing two variances is often applied prior to conducting a t-test on two means. The aim is to determine if the equal variance is violated.

Assume that the distribution of the population being sampled is normal, the chi-squared value for testing

2 =

Example 9 A manufacturer of cell phone batteries claims that the life of his batteries is approximately normally distributed with a standard deviation equal 1.05 years. If a random sample of 15 of these batteries has a standard deviation of 1.3 years, do you think that > 1.05 years? Use a 0.01 level of significance.

Solution: 1) H0: 2 = 1.1025H1: 2 > 1.1025

2) Let a = 0.01

3) Compute for Chi-Square Variance test.

2 = 21.46, P = 0.0904

4) Critical Region:

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2 = 2 is given by0

( n – 1 ) s 2 2

0


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2 > 29.14.

5) Do not reject H0. The 2 statistic is not significant at the 0.01 level. However, based on the P-value 0.09 there is evidence that > 1.05.

Goodness-of-Fit Test

Throughout this chapter we have been concerned with the testing of statistical hypotheses about a single population parameters such as , 2, and p. Now we shall consider a test to determine if a population has a specified theoretical distribution. The test is based on how good a fit we have between the frequency occurrence of observations in an observed sample and the expected frequencies obtained from the hypothesized distribution.

The observed frequencies will always differ from the expected frequencies due to sampling error. But are these differences significant? The chi-square goodness-of-fit test will enable one to determine the answer.

2 =

where; O = observed frequencies

E = expected frequenciesk – 1 degrees of freedom

Note: The expected frequency for each category must be at least 5.

Example 10 There are three gates of Juan G. Macaraeg National High School. The principal would like to know if the gates are equally utilized. As an experiment, 2700 students are observed as they enter the

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29.142

21.46

( O – E ) 2 E


__________________________________________________________________school. The number of students enter the gate in Canarvacanan were 1005, in Sta. Fe were 985, and in Sto. Niño were 710. At 0.05 significance level, can we conclude that there is no difference in the use of three gates?

Solution: 1. H0: Students have no gate preference.

H1: Students show a gate preference.

2. Let a = 0.05.

3. Compute for the chi-square goodness-of-fit test.

2 = 60.39

4. Critical Region

2 0.05, 2 = 5.99

5. Decision: Reject H0.Conclusion: There is a large difference between the set of

observed frequencies and set of expected frequencies.

Test for Independence ( Categorical Data)

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The chi-square test procedure can also be used to test the hypothesis of independence of two variables of classifications. Suppose we wish to determine if each person’s blood type and eye color are related in any way. To find whether two observed characteristics of a member of a population are independent, we will use Test of Independence.

Suppose we pick a sample size n and classify the data in a two-way table on the basis of two variables. Such a table for determining whether the distribution according to one variable is contingent on the distribution of the other is called a contingency table. A contingency table with r rows and c columns is referred to as an r c table ( “r c” is read “ r by c” ).

The formula for Test for Independence:

2 =

where the summation extends over all rc cells in the r c contingency table. If 2 > 2 with v = ( r – 1 )( c – 1 ) degrees of freedom, reject the null hypothesis of independence at the level of significance; otherwise, do not reject the null hypothesis.

Example 11 A researcher wishes to see whether the age of an individual is related to milk consumption. The data are classified below.

AgeMilk Consumption

Low Moderate High16 – 25 10 15 1626 – 35 11 13 2736 – 45 16 28 5

45 and above 8 9 7

At = 0.05, is there a relationship between milk consumption and age?

Solution: 1) H0: Milk consumption of a person does not depend on his age.

H1: Milk consumption of a person depends in his age.

2) Let = 0.05.

3) Compute for 2

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( O – E ) 2 E

a


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2 = 22.37

4) Critical Region:

2 0.05, 6 = 12.59

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Conclusion: Milk consumption of an individual depends in his age.

Name:_______________________________________ Score:______________Section:______________________________________ Date: ______________

Exercise 2.1

1. The hospital record shows that the mean weight of a newly born baby is 8.3 lbs, with the standard deviation of 0.6 lbs. A researcher takes a sample of 100 newly born babies and found to have a mean of 7.8 lbs. Test the claim at 0.01 level of significance.

2. A sociologist finds that in the Philippines, the mean number of years of education is 10 while the standard deviation is 1.8. In Region I, a random sample of 100 people and found out that the sample mean is 12 years. At the 0.01 level of significance, test the claim that the mean for Region I is the same as the mean of the country.

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3. A survey found that women over the age of 60 consume an average of 1870 calories a day. In order to see if the number of calories consumed by women over age 60 living in Binalonan is the same, the researcher sampled 200 women over the age of 60 and found the mean number of calories consumed was 1997. The standard deviation of the sample was 56 calories. At a = 0.05, can it be concluded that there is no difference between the number of calories consumed by the women over age 60?

4. A manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 80 months. A consumer protection agency that wants to check this claim took a random sample of 50 batteries such batteries and found that the mean life for this sample is 73.75 months with a standard deviation of 7 months. Using the 2.5% significance level, would you conclude that the mean life of these batteries is less than 80 months?

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5. A company claims that the mean weight per banana it ships is 180 grams with a standard deviation of 10 grams. Data generated from a sample of 70 bananas randomly selected from a shipment indicated a mean weight of 193.5 grams per banana. Is there sufficient evidence to reject the company’s claim? Use a = 0.01.

6. The treasurer of a certain university claims that the mean monthly salary of their college professor is P 37,750 with a standard deviation of P 3000. A researcher takes a random sample of 100 college professors why were found to have a mean monthly salary of P 34,375. Do the 100 college professors have higher salaries than the rest? Test the claim at a = 0.02 level of significance.

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Exercise 2.21. Past experience indicates that the time for high school juniors to complete

a standardized test is a normal random variable with a mean of 60 minutes. If a random sample of 15 high school juniors took an average of 66 minutes to complete this test with a standard deviation of 5 minutes, test the hypothesis at 0.01 level of significance that = 60 minutes against the alternative that < 60 minutes.

2. A recent survey stated that households received an average of 30 telephone calls per week. To test the claim, a researcher surveyed 15 households and found that the average number of calls was 38 and the standard deviation is 4. At = 0.01, can the claim be rejected?

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3. Test the hypothesis that the average content of a particular soft drink is 1 liter if the contents of a random sample of 10 bottles are 1.04, 0.97, 1.01, 1.05, 0.97, 0.98, 1.02, 1.05, 0.97, and 0.97 liters. Use a 0.01 level of significance and assume that the distribution of contents is normal.

4. The president of a certain tricycle operators and drivers claims that the average mileage of tricycles is less than 80000. A sample of 16 tricycles has an average mileage of 90000, with standard deviation of 8000. At = 0.02, is there enough evidence to reject the president’s claim?

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Exercise 2.3

1. A large automobile manufacturing company is trying to decide whether to buy brand X or brand Y tires for its new models. To help arrive at a decision, an experimented is conducted using 100 of each brand. The tires are run until they wear out. The results are

Brand Mean ( in kilometers) SD ( in kilometers )X 60,000 750Y 75,750 500

Test the hypothesis that there is no difference in the 2 brands of tires. Use a 0.03 level of significance.

2. Two types of plastic are suitable for use by an electric component manufacturer. The breaking strength of this plastic is important. If is known that psi. From a random sample of size n = 60 and m

= 72, we obtain psi and psi. The company will choose the plastic with larger breaking strength. Based from the sample, should the company choose plastic 1? Use 0.01 level of significance.

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3. A Mathematics department head claims that the average score of Special Science Class exceeds the average score of Regular class by at least 20 points in the progress test. To test his claim, he choose at random of 100 SSC students and found out that their average score was 80 and standard deviation of 9, while120 Regular class students had an average score of 67 and standard deviation of 7. Test the department head’s claim using a 0.01 level of significance.

4. The following data, recorded in kilometers per liter, represent the fuel consumption of two vehicles tested at 90-kilometer per hour steady-speedy tests;

Vehicle Sample ( n ) Mean Standard DeviationJeepney 67 27 2.3

Mini - Bus 59 29 4.1

Test the hypothesis that Mini buses, on the average, exceed similarly equipped jeepneys by 5 kilometers per liter. Use a 0.01 level of significance.

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Exercise 2.4

1. To find out whether a new drug will cure diabetes, 15 mice with an advanced stage of disease, are selected. Survival times, in years, from the time the experiment commenced are as follows:

Treatment 8.9 6.8 6.2 2.9 7.9 5.8 4.2 6.1No treatment 2.2 2.3 2.5 3.6 2.4 3.1 2.1

At the 0.05 level of significance, can the drug be said be effective?

2. Teacher X conducted a review class in his Chemistry class. He gave a test before and after the review and gathered the following data:

STUDENT1 2 3 4 5 6 7 8 9

Score Before Review

15 19 19 17 24 29 26 11 18

Score After Review

17 29 41 20 32 28 39 30 27

At = 0.01 level of significance, is the review class effective?

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3. It is claimed that a new diet will reduce a person’s weight by 3 kilograms on the average in a period of 10 days. The weights of 8 women who followed this diet were recorded before and after a 20-day period.

WOMAN1 2 3 4 5 6 7 8

Weight before

57.6 60.5 89.8 67.9 57.6 54.2 68.3 69.3

Weight after

50.6 60.3 62.4 60.2 51.6 55 59.4 60.3

Test the claim at = 0.02 level of significance.

4. Twenty subjects were used in an experiment to determine if an atmosphere involving exposure to carbon monoxide has an impact on breathing capability. The subjects were exposed to breathing chambers, one of which contained a high concentration of CO. The average breath-taken-per-minute without CO is 50 and standard deviation of 2, while the average breath-taken-per-minute exposed with CO is 49 and the standard deviation of 6. Test the hypothesis that the mean breathing frequency is the same for the two environments. Use a 0.01 level of significance.

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Exercise 2.5

1. It is claimed that a certain drug causes a successful cure in 70% of all cases. At a 0.01 level of significance, would you agree to this claim if in a sample of 800 cases 550 were cured?

2. At a certain college it is estimated that at most 50% of the students ride motorcycles to class. Does this seem to be a valid estimate if, in a random of 100 college students, 40 are found to ride motorcycles to class? Use a 0.01 level of significance.

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3. A telephone company claims that two-third of the homes in a certain city have landline telephones. Do we have reason to doubt this claim if, in a random sample of 1000 homes in this city, it is found that 878 have telephones. Use a 0.05 level of significance.

4. Supposed that, in the past, 40% of all adults favored death penalty. Do we have reason to believe that the proportion of adults favoring death penalty today has increased if, in a random sample of 40 adults, 19 favor death penalty? Use a 0.02 level of significance.

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Exercise 2.6

1. An urban community would like to show that the incidence of breast cancer is higher than in a nearby rural area. If it is found that 250 of 3000 adult women in the urban community have breast cancer and 105 of 1000 adult women in the rural community have breast cancer, can we conclude that breast cancer is more prevalent in the urban community? Use a 0.0l level of significance.

2. A study was made to determine whether more Filipinos than Italians prefer white champagne to pink champagne at weddings. Of the 1000 Filipinos selected at random, 178 preferred white champagne, and of the 3000 Italians selected, 80 preferred white champagne. Can we conclude that a higher proportion of Filipinos than Italians prefer white champagne at weddings? Use a 0.02 level of significance.

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Exercise 2.7

1. A soft-drink dispensing machine is said to be out of control if the variance of the contents exceeds 1.2 deciliters. If a random sample of 200 drinks from this machine has a variance 0f 1.69, does this indicate at the 0.01 level of significance that the machine is out of control?

2. A company claims that the variance of the sugar content of its ice cream is equal to 50. A sample of 200 servings is selected, and the sugar contents are measured and found out that the sample variance is 40. At = 0.05, is there sufficient evidence to believe the claim?

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3. A medical researcher believes that the standard deviation of the temperature of newborn babies is equal to 0.70. A sample of 50 infants was found to have a standard deviation of 0.750. Assume that the variable is normally distributed; does the evidence support the medical researcher’s claim? Use a 0.01 level of significance.

4. A study is conducted to compare the length of time between men and women to encode a certain phrase in their cellular phones. The data are tabulated below:

Men Womenn 10 23sd 1.2 3.7

Test the hypothesis against the alternative hypothesis that variance of men is greater than of the women. Use a 0.01 level of significance.

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Exercise 2.8

1. A machine is supposed to mix peanuts, cashews, hazelnuts, and pecans in the ratio 4: 3: 2: 1. A can containing 640 of these mixed nuts was found to have 250 peanuts, 190 cashews, 160 hazelnuts, and 40 pecans. At a 0.05 level of significance, test the hypothesis that the machine is mixing the nuts in the ratio 4: 3: 2: 1.

2. A supervisor at a certain cinema wants to determine if there is any preference in the flavors of popcorn that were sold in a day. A random sample of sales is selected, and the data are shown below. At = 0.05, are the flavors selected with equal frequency?

Flavor Butter Barbeque Cheese Plainsold 38 51 36 45

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Exercise 2.9

1. The department head of Mathematics wanted to determine if there are significant differences in the way his instructors handled out pass or fail grades. He set a 0.05 level of significance. The data are shown below:

Instructor PASS FAILX 50 10Y 61 3Z 27 5

2. A sociology study compared 3 groups in their responses to 1 question: “Are you happier now than you were 4 years ago?” Their responses are tabulated below

GROUPRESPONSE

More Less SameProfessional 59 29 19Blue collar 16 12 20

Unskilled laborers 17 57 5

Is there significant difference among their responses? Use a 0.01 level of significance.

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3. In a study of car accidents and drivers who use cellular phones, the following sample data are obtained. At a = 0.01, test the claim that the occurrence of accidents in independent of the use of cellular phones.

Had accident Had no accidentCellular phone user 100 300

Non-cellular phone user 15 400

4. The supermarket sells red and white eggs in sizes small, medium, large, and extra large. The table shows the number of cartons sold for the various sizes and colors during a one-month period.

TYPE OF EGG

EGG SIZESmall Medium Large Extra Large

Red 7000 4077 5011 2080White 4810 5500 8203 4700

Is egg color preference dependent on the size purchased? Test at 0.01 level of significance.

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Chapter 2 Hypothesis Testing

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Transcript of Chapter 2 Hypothesis Testing