CHAPTER 3 HYPOTHESIS TESTING


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1 CAPTER 3 YPOTESIS TESTING Expected Outcomes Able to a population mean when population variance is known or unknown. Able to the difference between two populations mean when population variances are known or unknown. Able to paired data using z and t. Able to population proportion using z. Able to the difference between two populations proportion using z. Able to a population variance and the difference between two populations variances. Able to determine the relationship between hypothesis ing and confidence interval. Able to solve hypothesis ing using Microsoft Excel. PREPARED BY: DR SITI ZANARIA SATARI & SITI ROSLINDAR YAZIZ
2 CONTENT 3.1 Introduction to ypothesis Testing 3. Test ypothesis for Population Mean with known and unknown Population Variance 3.3 Test ypothesis for the Difference Population Means with known and unknown Population Variance 3.4 Test ypotheses for Paired Data 3.5 Test ypotheses for Population Proportion 3.6 Test ypotheses for the Difference between Two Population Proportions 3.7 Test ypotheses for Population Variance 3.8 Test ypotheses for the Ratio of Two Population Variances 3.9 PValues in ypothesis Test 3.1 Relationship between ypothesis Tests and Confidence Interval
3 3.1 INTRODUCTION TO YPOTESIS TESTING A statistical hypothesis is a statement or conjecture or assertion concerning a parameter or parameters of one or more populations. Many problems in science and engineering require that we need to decide either to accept or reject a statement about some parameter, which is a decisionmaking process for evaluating claims or statement about the population(s). The decisionmaking procedure about the hypothesis is called hypothesis ing. 3 Methods of ypothesis Testing The traditional method The P  value method The confidence interval method
4 3.1.1 TERMS AND DEFINITION Definition 1a: A null hypothesis, denoted by is a statistical hypothesis that states an assertion about one or more population parameters. Definition 1b: The alternative hypothesis denoted by is a statistical hypothesis that states the assertion of all situations that not covered by the null hypothesis. : : : TWO TAILED TEST RIGT TAILED TEST LEFT TAILED TEST : 1 : 1 : 1 parameter A value
5 Types Of ypothesis Type of ypothesis Twotailed Onetailed Righttailed Lefttailed ypothesis : : 1 : : 1 : : 1 Note: (i) The should have equals sign and 1 should not have equals sign. (ii) The is on trial and always initially assumed to be true. (iii) Accept if the sample data are consistent with the null hypothesis. (iv) Reject if the sample data are inconsistent with the null hypothesis, and accept the alternative hypothesis.
6 Definition : A statistic is a sample statistic computed from the data obtained by random sampling. Z t,,, f Definition 3: The rejection (critical) region α, is the set of values for the statistics that leads to rejection of the null hypothesis. Definition 4: The acceptance region, 1 α is the set of values for the statistics that leads to acceptance of the null hypothesis. Definition 5: The critical value(s) is the value(s) of boundary that separate the rejection and acceptance regions. Definition 6: The decision rule of a statistical hypothesis is a rule that specifies the conditions under which the null hypothesis may be rejected. Reject if statistics > critical value 6
7 Type of Test ypothesis Rejection Region Graphical Display (ypothesis using statistic z with.5 ) Twotailed : : 1 Both sides Onetailed Righttailed Lefttailed : : 1 : : 1 Right side Left side 7
8 Definition 7: Rejecting the null hypothesis when it is true is defined as Type I error. P(Type I error) = (significance level) Definition 8: Failing to reject the null hypothesis when it is false in state of nature is defined as Type II error. P(Type II error) = Possible Outcomes: State of Nature Statistical Conclusion/decision Reject Not to reject is true Type I error Correct decision is false Correct decision Type II error
9 Example The additive might not significantly increase the lifetimes of automobile batteries in the population, but it might increase the lifetime of the batteries in the sample. In this case, would be rejected when it was really true, which committing a type I error. While, the additive might not work on the batteries selected for the sample, but if it were to be used in the general population of batteries, it might significantly increase their lifetime. ence based on the information obtained from the sample, would not reject the, thus committing a type II error.
10 ypothesis ing common phrase Is greater than Is above Is higher than Is longer than Is bigger than Is increased : 1 : Is greater than or equal Is at least Is not less than Is equal to Is exactly the same as as not changed from Is the same as Is less than Is below Is lower than Is shorter than Is smaller than Is decreased or reduced from : Is less than or equal Is at most Is not more than : : 1 : 1 Is not equal to Is different from as changed from Is not the same as
11 3..1 PROCEDURES OF YPOTESIS TESTING Step 1: Formulate a hypothesis and state the claim Twotailed : : 1 OR Step : Choose the appropriate statistic, and calculate the sample statistic value: Z Righttailed t,,, f : : 1 Step 3: Establish the criterion by determining the critical value (point) and critical region Significance level value, Inequality (, >, <) used in the OR Step 4: Make a decision to reject or not to reject the. 1 Step 5: Draw a conclusion to reject or to accept the claim or statement. Lefttailed : : 1
12 ypothesis Testing: Step by Step Extract all given information Define Parameter Define and 1 Choose appropriate statistics Find critical value Test the hypothesis (rejection region) Make a conclusion there is enough evidence to reject/accept the claim at α
13 3.: TEST YPOTESES FOR POPULATION MEAN, μ WIT KNOWN AND UNKNOWN POPULATION VARIANCE Twotailed : : 1 Righttailed : : 1 Lefttailed : : 1
14 Test Statistics of ypothesis Testing for Mean μ z X n z X t s n X s n Where: population mean NOTE: Z and t are statistics
15 The Rejection Criteria (1) i. If the population variance, is known, the statistic to be used is z x z. / n ~ Therefore the rejection procedure for each type of hypothesis can be summarised as in following Table 3.4. Table 3.4: ypothesis ing for μ with known σ 1 Statistical Test Reject if : 1: z z or z z x z : 1: z z / n : : 1 z z
16 The Rejection Criteria () ii. If the population variance, is unknown and the sample size is large, i.e. n 3, then the statistic to be used is z x z. s/ n ~ Therefore the rejection procedure for each type of hypothesis can be summarised as in following Table 3.5. Table 3.5: ypothesis ing for μ with unknown σ and n 3 1 Statistical Test Reject if : 1: z z or z z x z : 1: z z s/ n : : 1 z z
17 The Rejection Criteria (3) iii. If the population variance, is unknown and the sample size is small, i.e. n 3, then the statistic to be used is x t ~ t, v where v n 1 s/ n Therefore the rejection procedure for each type of hypothesis can be summarised as in following Table 3.6. Table 3.6: ypothesis ing for μ with unknown σ and n 3 1 Statistical Test Reject if : 1: t t or t t x : 1: t t t, n 1 s/ n : : t, n 1, n1 1 t, n1
18 Example 3 Most watertreatment facilities monitor the quality of their drinking water on an hourly basis. One variable monitored is p, which measures the degree of alkalinity or acidity in the water. A p below 7. is acidic, above 7. is alkaline and 7. is neutral. One watertreatment plant has target a p of 8.5 (most try to maintain a slightly alkaline level). The mean and standard deviation of 1 hour s results based on 31 water samples at this plant are 8.4 and.16 respectively. Does this sample provide sufficient evidence that the mean p level in the water differs from 8.5? Use a.5 level of significance. Assume that the population is approximately normally distributed. Solution: Step 1: Formulate a hypothesis and state the claim. X: p level in the water 1 : 8.5 : 8.5 claim
19 Example 3: solution Step : Choose the appropriate statistic and calculate the sample statistic value. Since is unknown, i.e..16 s and n 3, x the statistic is z s/ n Step 3: Establish the criterion by determining the critical value and rejection region. Statistical Test 1 Reject if : 1: z z or z z x z : 1: z z s/ n : : 1 z z 19
20 Example 3: solution Step 3: Establish the criterion by determining the critical value and rejection region. 1 Statistical Test Reject if : : x 1 z o z z or z z s/ n Given.5 and the is twotailed, hence the critical values are z and z Step 4: Make a decision to reject or fail to reject the. Since z z,.5 then we reject. Step 5: Draw a conclusion to reject or to accept the claim or statement. At.5, the sample provide sufficient evidence that the mean p level in the water differs from 8.5.
21 3.3 TEST YPOTESES FOR TE DIFFERENCE BETWEEN TWO POPULATIONS MEAN : 1 : 1 1 : 1 : 1 1 : 1 : 1 1 Twotailed Righttailed Lefttailed
22 Test Statistics for the Difference between Means z x x 1 n n 1 1 o z z x s 1 p x 1 1 n n 1 s p o z t x 1 o, n n s p x 1 1 n n 1 ( n 1) s ( n 1) s n n t 1 z x 1 x s n s n 1 1 o z t x 1 x s n s n 1 1 s1 s n1 n s 1 s n1 n n 1 n 1 1 o t,
23 Example 4 The overall distance travelled of a golf ball is ed by hitting the ball with the golf stick. Ten balls selected randomly from two different brands are ed and the overall distance is measured and the data is given as follows. Overall distance travelled of golf ball (in meters) Brand Brand By assuming that both population variances are unequal, can we say that both brands of ball have similar average overall distance? Use α =.5.
24 Step 1: X 1 : Overall distance travelled of golf ball from brand 1 X : Overall distance travelled of golf ball from brand Example 4: solution The hypothesis is : ( claim) 1 : 1 1 Step : Statistic Brand 1 Brand n 1 1 x s Since 1 and are unknown,, and n13, n 3, then the statistic is 1 t ( x x ) ( ) 1 s1 s n n
25 Step 3: Given.5 and the is twotailed. The critical value is t t.5, 17, ν Example 4: solution s s n1 n 1 1 where ν s 1 s n1 n 1 1 n 1 n Step 4: Since t t.5, , is rejected. Step 5: At.5, there is no significant evidence to support that both brands of ball have similar average overall distance
26 3.4 TEST YPOTESES FOR PAIRED DATA Twotailed : D : 1 D Righttailed : D : 1 D Lefttailed : D : 1 D is population mean difference, D 1 where t Test Statistics x s D D D / n ~ t, v D X1X differences between the paired sample n is number of paired sample x, s D D, are the mean and standard deviation for the difference of paired sample, respectively vn1, degrees of freedom
27 Example 5 A new gadget is installed to air conditioner unit(s) in a factory to minimize the number of bacteria floating in the air. The number of bacteria floating in the air before and after the installation for a week in the factory is recorded as follows. Before After Is it wise for the factory management to install the new gadget? By assuming the data is approximately normally distributed, the hypothesis at 5% level of significance.
28 1 : D : (wise to install the new gadget) D where D X 1 BeforeAfter Before, After, X D X X x s t D D Example 5: solution / 7 t.5, ,6 Since t.881 t is rejected. It is wise for the factory management to install the new gadget, at 5% level of significance.
29 3.5 TEST YPOTESES FOR POPULATION PROPORTION The hypothesis: Twotailed Righttailed Lefttailed : : 1 OR : : 1 OR : : 1 The Test Statistics: z p 1 n ~ z where p x n  sample proportion  given population proportion
30 Example 6 An attorney claims that at least 5% of all lawyers advertise. A sample of lawyers in a certain city showed that 63 had used some form of advertising. At α =.5, is there enough evidence to support the attorney s claim? Step 1: X is the number of lawyers advertise 1 :.5 :.5 claim
31 Example 6: solution Step : Since n and x 63, then 63 p.315. The statistic is z Step 3: Given.5 and the is lefttailed, hence the critical value is z Step 4: Since z z, then we accept..5 Step 5: At.5, there is enough evidence to support the attorney s claim.
32 3.6 TEST YPOTESES FOR DIFFERENCE BETWEEN TWO POPULATIONS PROPORTION The hypothesis: Type of Test ypothesis Decision on Rejection : 1 Twotailed Reject if z z or z z : Righttailed Lefttailed The Test Statistics: 1 1 : 1 : 1 1 : 1 : 1 1 Reject if z Reject if z z z If : z p1 p n n 1 ~ z If : z p p 1 p p p X X 1 1 ~z where pp n1 n p 1 1 n1 n 3
33 Example 7 An experiment was conducted in order to determine whether the increased levels of carbon dioxide (CO) will kill the leafeating insects. Two containers, labeled X and Y were filled with two levels of CO. Container Y had double of CO level compared to container X. Assume that 8 insect larvae were placed at random in each container. After two days, the percentage of larvae that died in container X and Y were five percent and ten percent, respectively. Do these experimental results demonstrate that an increased level of CO is effective in killing leafeating insects larvae? Test at 1% significance level.
34 Example 7: solution Step 1: X: the number of the number of larvae that died in container X Y: the number of the number of larvae that died in container Y 1 : Y X : ( claim) Y X Step : Statistic Y X n 8 8 p.1.5 x 8 4 The statistic is z ( p.1.5 Y px) P p Pp n 8 8 Y nx xy xx 84 where Pp.75 n n 8 8 Y X
35 Example 7: solution Step 3: Given.1 and the is righttailed, hence the critical value is z Step 4: Since z z , then we failed to reject Step 5: At.1, there is no significant evidence to support that an increased level of carbon dioxide is effective in killing higher percentage of leafeating insects larvae..
36 3.7 TEST YPOTESES FOR A POPULATION VARIANCE The hypothesis: Type of Test ypothesis Twotailed : 1: Righttailed : 1: Lefttailed : 1: The Test Statistics: Decision on Rejection Reject if 1, 1, 1, n 1 1, n 1 or n1 s ~ s, vn1 is the sample variance, is the given variance 36
37 Example 8 Listed below are waiting times (in minutes) of customers at a bank The management will open more teller windows if the standard deviation of waiting times (in minutes) is at least.9 minutes. Is there enough evidence to open more teller windows at α =.1?
38 Example 8: solution Step 1: X is waiting times (in minutes) of customers at a bank :.9 minutes (open more teller windows) 1 :.9 minutes Step : n 6 customers x 7.13 minutes s.43 minutes The statistic is n s Step 3: Given.1 and the is lefttailed, hence the critical value is ,5 Step 4: Since.99, , then we failed to reject. Step 5: At.1, there is enough evidence to open more teller windows.
39 3.8 TEST YPOTESES FOR TE RATIO OF TWO POPULATION VARIANCES The hypothesis: Type of Test ypothesis Decision on Rejection Twotailed : 1 1 : 1 Reject where f if f f or f f 1, n1 1, n1 f 1, n1 1, n 1, n1 1, n 1 1, n1, n11 Righttailed : 1 1 : 1 Reject if f f, n1 1, n 1 Lefttailed : 1 1 : 1 Reject if where f1, n1 1, n 1 f f 1, n1 1, n 1 f 1, n1, n11 The Test Statistics: s f ~ fv 1, v where v 1 n1 1, v n 1 s 1 39
40 Example 9 A manager of computer operations of a large company wants to study the computer usage of two departments within the company. The departments are uman Resource Department and Research Department. The processing time (in seconds) for each job is recorded as follows: uman Resource Research Is there any difference in the variability of processing times for the two departments at α =.5.
41 Example 9: solution Step 1: X 1 : processing time (in seconds) for each jobs from for uman Resource Department X : processing time (in seconds) for each jobs from for Research Department : 1 : 1 1 claim Step : n1 5, x1 7.8, s1 3.3 n 6, x 8.5, s The statistic is F s s Step 3: Given.5 and the is twotailed, hence the critical value are F, n11, n1 F 1, n11, n1 F ,4, F.975,4,5.168 F F , n1, n11.5,5,4
42 Example 9: solution Step 4: Since F.975,4,5 f F.5,4,5 reject , then we failed to Step 5: At.5, there is no difference in the variability of processing times for the two departments.
43 3.9 PValues IN YPOTESIS TESTING The Pvalue (Probability value) is the smallest level of significance that would lead to rejection of the null hypothesis with the given data Finding the Pvalue Statistical Table Step 1: Find the area under the standard normal distribution curve corresponding to the z value. Step : Subtracting the area from.5 to get the Pvalue for a righttailed or lefttailed. To get the Pvalue for a twotailed, double the area after subtracting. Calculator (Casio fx57 MS) Step 1: Find the area under the standard normal distribution curve corresponding to the z value. Step : The area obtained is the Pvalue for a righttailed or lefttailed. To get the Pvalue for a twotailed, double the area. Pvalue P Z R
44 Procedures of ypothesis Testing using PValue Approach Step 1: Formulate a hypothesis and state the claim Twotailed : : 1 Step : Choose the appropriate statistic, and calculate the sample statistic value. Step 3: Find the Pvalue Righttailed : : 1 Step 4: Make a decision to reject or not to reject the. If P value Reject If P value Do not Reject Step 5: Draw a conclusion to reject or to accept the claim or statement. Lefttailed : : 1 44
45 Example 1 Most watertreatment facilities monitor the quality of their drinking water on an hourly basis. One variable monitored is p, which measures the degree of alkalinity or acidity in the water. A p below 7. is acidic, above 7. is alkaline and 7. is neutral. One watertreatment plant has target a p of 8.5 (most try to maintain a slightly alkaline level). The mean and standard deviation of 1 hour s results based on 31 water samples at this plant are 8.4 and.16 respectively. Does this sample provide sufficient evidence that the mean p level in the water differs from 8.5? Use a.5 level of significance. Assume that the population is approximately normally distributed. [Example 3] Solve this problem using Pvalue approach.
46 Example 11: solution
47 Example 11: solution
48 Pvalue Using Excel Test For Mean Step 1: Click Menu Data Data Analysis Descriptive Statistics click OK Step : a) The commands for t are (i) t = (Mean  )/Standard Error (ii) Pvalue for a twotailed = T.DIST.T(ABS(t), degrees of freedom) Pvalue for a righttailed = T.DIST.RT((ABS(t), degrees of freedom) Pvalue for a lefttailed = T.DIST(ABS(t), degrees of freedom, 1) Note: Standard Error is a standard deviation divided by the square root of the number of data which can be written as s.e.. n
49 Example 11 A petroleum company is studying to buy an additive for improving the distilled product. The company estimates the cost of the additive, which is RM1 million for 5 tonnes. Ten consultant companies submitted their tenders with the following estimates (in million RM): Do you think the petroleum company over estimates the cost of the additive? Give your reason. Use Pvalue method.
50 Example 11: solution Step 1: Formulate the hypothesis 1 : 1 C : 1 (claim: company over estimate the cost) C Step : Key in the data, select data data analysis Descriptive Statistics click OK
51 Example 11: solution Output from Excel: Column1 Mean Standard Error Median 1.1 Mode.97 Standard Deviation Sample Variance Kurtosis Skewness Range.75 Minimum.95 Maximum 1.7 Sum Count 1 Confidence Level(95.%) The values highlighted will be used to calculate t t and Pvalue are calculated using Excel command as follows: t = ( )/ Since the case is t and lefttailed, Pvalue = T.DIST( ,9,1) t Pvalue Step 3: Pvalue.976 Step 4: Since Pvalue.976.5, then we do not reject. Step 5: At.5, there is not enough evidence to support the claim that the petroleum company over estimate the cost of the additive.
52 Pvalue Using Excel Test For Difference Mean Step 1: Test the difference in variability > F.TEST(data set 1, data set ) Step : Click Menu Data> Data Analysis> Choose the appropriate (i.e.: ttest: TwoSample Assuming Unequal Variances)> click ok Step 3: Variable 1 range> select the data set 1 Variable range> select the data set ypothesized mean difference> value of μ Alpha> value of significance level, α Step 4: Pvalue for a twotailed = P(T<=t) twotails (depends on distribution used) Pvalue for a righttailed = P(T<=t) onetail (depends on distribution used) Pvalue for a lefttailed = 1 P(T<=t) onetail (depends on distribution used)
53 Example 1 A company is considering installing a new machine to assemble its product. The company is considering two types of machine, Machine A and Machine B but it will by only one machine. The company will install Machine B if the mean time taken to assemble a unit of the product is less than Machine A. Table below shows the time taken (in minutes) to assemble one unit of the product on each type of machine. Machine A Machine B At 1% significance level, the difference in variability between the two types of machines. Which machine should be installed by the company to assemble its product?
54 Example 1: solution Step 1: Formulate the hypothesis 1 : A B : (claim) A B and 1 : : A B A B Pvalue =.39 >.1 Thus, Failed to reject. There is no difference in the variability.
55 Example 1: solution Step : Key in the data in Excel and choose the ttest: TwoSample Assuming equal Variances ttest: TwoSample Assuming Equal Variances machine A Machine B Mean Variance Observations 8 8 Pooled Variance ypothesized Mean Difference df 14 t Stat P(T<=t) onetail.1417 t Critical onetail P(T<=t) twotail.4854 t Critical twotail Step 3: The is onetailed, hence Pvalue =.14 Step 4: Since P value.14.1, then we do not reject. Step 5: At 1% significance level, machine A should be installed.
56 Pvalue Using Excel Test For Paired Data Step 1: Click Menu Data> Data Analysis> Choose the appropriate (i.e.: ttest: Paired Two Sample for Means)> click ok Step : Variable 1 range> select the data set 1 Variable range> select the data set ypothesized mean difference> value of μ Alpha> value of significance level, α Step 3: Pvalue for a twotailed = P(T<=t) twotails (depends on distribution used) Pvalue for a righttailed = P(T<=t) onetail (depends on distribution used) Pvalue for a lefttailed = 1 P(T<=t) onetail (depends on distribution used)
57 1 : D Example 13: refer data example 5 : (wise to install the new gadget) D Pvalue =.15 <.5 Thus, reject. At 5% significance level, it is wise to install the new gadget.
58 3.1 RELATIONSIP BETWEEN YPOTESIS TEST & CONFIDENCE INTERVAL There is a relationship between the confidence interval and hypothesis about the parameter,. Let say ab, is a 1 1% confidence interval for the, the of the size of the hypothesis : : 1 will lead to rejection of if and only if is not in the 1 1% confidence interval ab,. Notes: This relationship should be checked for twotailed only.
59 Example 14 By considering Example 3.3 again, the 95% confidence interval for is 8.4 z , Since 8.5 is not included in this interval, the is rejected. So, the decision making or conclusion is the same as in Example 3.3 and Example 3.1.
60 REFERENCES 1. Montgomery D. C. & Runger G. C. 11. Applied Statistics and Probability for Engineers. 5 th Edition. New York: John Wiley & Sons, Inc.. Walpole R.E., Myers R.., Myers S.L. & Ye K. 11. Probability and Statistics for Engineers and Scientists. 9 th Edition. New Jersey: Prentice all. 3. Navidi W. 11. Statistics for Engineers and Scientists. 3 rd Edition. New York: McGrawill. 4. Bluman A.G. 9. Elementary Statistics: A Step by Step Approach. 7 th Edition. New York: McGraw ill. 5. Triola, M.F. 6. Elementary Statistics.1 th Edition. UK: Pearson Education. 6. Satari S. Z. et al. Applied Statistics Module New Version. 15. Penerbit UMP. Internal used. Thank You NEXT: CAPTER 4 ANALYSIS OF VARIANCE
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