CHAPTER 5 STATISTICAL INFERENCE. 1.0 Hypothesis Testing. 2.0 Decision Errors. 3.0 How a Hypothesis is Tested. 4.0 Test for Goodness of Fit

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1 Chater 5 Statistical Inference 69 CHAPTER 5 STATISTICAL INFERENCE.0 Hyothesis Testing.0 Decision Errors 3.0 How a Hyothesis is Tested 4.0 Test for Goodness of Fit 5.0 Inferences about Two Means It ain't easy being green. Kermit

2 Chater 5 Statistical Inference 7 STATISTICAL INFERENCE.0 HYPOTHESIS TESTING A hyothesis is a statement, inference or tentative exlanation about a oulation that can be tested by further investigation. A hyothesis test is a statistical test that assists in the decision to rove or disrove the statement. In most cases, it may be easier to disrove a hyothesis than to verify it. Samle measurements are taken and inferences made about the total oulation based on the samle statistics. The inferences and the resulting decisions could be correct or incorrect. In hyothesis testing as well as all other statistical analyses, the risks of making a wrong decision are redetermined.. Null and Alternate Hyotheses The symbol µ reresents the oulation or universe mean and x reresents the samle mean. The symbol σ reresents the oulation standard deviation and σ or s reresent the standard deviation calculated from the samle. The oulation values µ and σ are arameters and the samle values x and σ or s are statistics. The following hyothesis is to be tested: The number 5 wheel bearings made by the Canae Bearing Comany have an average diameter of 0.65 inches. Based on the samle statistics, is this statement true or false? The objective of hyothesis testing is to verify the validity of the statement. The null hyothesis is the hyothesis to be roven or disroven. It is reresented by H 0. The alternate hyothesis is the hyothesis that is acceted if the null hyothesis is rejected by the test. It is reresented by H. For the test, the hyotheses are stated as follows: Null Hyothesis, H 0 : µ = 0.65 Alternate Hyothesis, H : µ Tyes of Hyothesis Tests Hyotheses concerning both measurements (variables or continuous data) or counts (attribute or discrete data) may be tested. The hyothesis H 0 : µ = 0.65 is an examle using continuous data. A hyothesis concerning discrete data may be stated as follows: The oulation from which a samle of one hundred arts was taken has a rocess average of % defective. H 0 : =.0 or H 0 : n = 00 x.0 =

3 QReview 7 The symbol reresents the oulation mean and reresents the samle mean. The symbol n reresents the average number of defective arts in the samle for both the hyergeometric and the binomial distributions. For the Poisson distribution, n is the average number of defects in the samle. In hyothesis testing, a general rule is to use the normal curve areas (Z - statistics) when the samle size n is thirty or more. When n is less than thirty, the t distribution areas (t - statistics) and corresonding tables are used. When n is very large (n ), the Z table and t table are virtually identical. The method of testing a hyothesis is exactly the same; the only difference is the table that is used. In the revious chater n - was used instead of n in the denominator of the variance formula. Division by n - is called division by degrees of freedom. The term degrees of freedom is so named because only n - linear comarisons can be made among n choices. If there are three job candidates and three jobs to fill, there are three choices for the first job. Once it is filled, there are two choices for the second job. After the second job is filled, there is only one choice for the third job. The third job can only be filled by the remaining erson, so there are n - or two degrees of freedom. To use the t table, the number of degrees of freedom is required..0 DECISION ERRORS Anytime samle data are used to make decisions or inferences about the entire oulation, decision errors can be made. In hyothesis testing and accetance samling, two tyes of errors can occur. Tye I error, alha ( α ), rejection of a true null hyothesis Tye II error, beta ( β ), accetance of a false null hyothesis Null hyothesis (H o ) is true Null hyothesis (H o ) is false Do not reject H o no error ( - α) Tye II error (β) Reject H o Tye I error (α) no error ( - β) The level of significance is denoted by α. This is the robability of rejecting a null hyothesis when it is true or the robability of making a wrong decision. The confidence interval is denoted by -. The confidence interval is the robability of not rejecting a null hyothesis when it is true or the robability of making a right decision. The robability of not rejecting a null hyothesis when it is false is β. This is the robability of making a wrong decision. The term - β is referred to as the ower of test and is the robability of rejecting a null hyothesis when it is false. This is the robability of making a right decision when the null hyothesis is truly false. In accetance samling, α and β are referred to as the roducer s risk and consumer s risk resectively.

4 Chater 5 Statistical Inference HOW A HYPOTHESIS IS TESTED 3. One or two tailed test? The alternate hyothesis determines whether the test will be a one tailed test or a two tailed test. If H is stated as H : µ.65, then the test is a two tailed test. If H is stated as H : µ <.65 or H : µ >.65, the test is a one tailed test. Sketch for a two tailed test (α =.05) Alternate Hyothesis, H : µ.65 (Area =.05) (Area =.05) Z Sketches for a one tailed test ( α =.05) Alternate Hyothesis Alternate Hyothesis H : µ <.65 H : µ >. 65 (Area =.05) (Area =.05) Z Z 3. Examles Examle Examle is a ste by ste rocedure on how to erform a hyothesis test. A samle of 00 bearings is taken. The diameter is measured and recorded for each art in the samle. The mean and standard deviation are calculated. The following results are obtained: x = 0.67 inches, s = 0.08 inches, n = 00

5 QReview 74 The distribution of individual data values has a mean of 0.67" and a standard deviation of 0.08". Since a samle mean is to be tested against a oulation mean, the test distribution is a distribution of averages. The distribution of averages has a mean of 0.67" and a standard deviation or standard error of n. Could the oulation from which this samle was selected have an average diameter of 0.65 inches? Test this hyothesis against a level of significance of.05 ( α =.05). Ste ) State the null and alternate hyotheses. H 0 : µ = 0.65 H ; µ 0.65 Ste ) Draw a sketch showing µ =.65 and x =.67 on the x scale x Ste 3) Determine the critical values of Z and the region of rejection. Since H : µ.65, this will be a two tailed test. Half of α will define the rejection region on the left tail of the curve and half of α will define the rejection region on the right tail of the curve. Half of α is.05/ =.05. The samle size is greater than thirty, therefore the correct table to use is the standard normal curve or Z table. Either draw another sketch or suerimose a Z scale on the first sketch. Look u =.475 in the table. The corresonding Z values are ±.96. These are the critical values. The critical values are the Z values that searate the rejection region from the accetance region. On the sketch, shade in the areas to the left of -.96 and to the right of The shaded areas indicate the regions of rejection. The area of the accetance region is.95, therefore this test will have a confidence interval of.95.

6 Chater 5 Statistical Inference 75 (Area = ½ α =.05) (Area = ½ α =.05) x Z Ste 4) Calculate the value of Z that corresonds with x using the x to Z conversion formula. Z = Z = ( x µ ) ( ) = s. 08 n 00 ( x µ ) s n = = = If the calculated value of Z is between the critical values, it falls in the accetance region and the hyothesis is not rejected. If the calculated value of Z is to the left of -.96 or to the right of +.96, the hyothesis that µ =.65 is rejected in favor of the alternate hyothesis that µ.65. Ste 5) Make a decision to reject or not reject the null hyothesis. The calculated Z value of +.5 lies in the rejection region, therefore H 0 is rejected in favor of H. It is concluded that µ.65. Examle Given x = 30.5, s =.4, and n = 0. Test the hyothesis µ = 30 against the alternative µ > 30 at a level of significance of.05. This will be a one tailed test because the alternate hyothesis is µ > 30. Since n = 0, the t table is used to determine the critical value and rejection region.

7 QReview 76 The calculated value of t is t = ( x µ ) ( ). 5 = = = + 6. s n x (α =.05) t The critical value is t = The area to the right of the critical value is the region of rejection. The calculated value t =.6 falls in the accetance region, therefore H 0 is not rejected. It can be concluded that the samle could have reasonably come from a oulation whose mean is thirty ( µ = 30). Examle 3 A samle of fifty arts is taken from a large shiment of a urchased roduct. The arts are insected and categorized as either defective or non-defective. The insector finds two defective arts. ( = /50 =.04 or 4% defective). The sulier states that his rocess yields 3% defective roduct. Could the sulier be telling the truth? Test at a level of significance of.0. This roblem can be worked out by stating the null hyothesis in terms of or in terms of n.

8 Chater 5 Statistical Inference 77 The standard deviation is modified deending on which null hyothesis is used. The Z value and the resulting decision will be the same. The formulas below are from chater 4, age 60. The standard deviation in terms of n for the binomial distribution = nq q ( ) The standard deviation in terms of for the binomial distribution = = n n The hyothesis test in terms of is H o : =.03, H : >.03, =.04 σ = ( ). 03(. 97) ( ) ( ) = =. 04 Z = = =. 4 n 50 σ. 04 The hyothesis test in terms of n is H o : n =.5, H >.5, n = σ n ( n n) ( 5. ) = n( ) = ( 50)(. 03)(. 97) = 06. Z = = =. 4 σ 06. n n (Area =.0) Z The critical value of Z is The calculated value of Z is +.4. This value lies in the accetance region, therefore H 0 is not rejected. The test suorts the sulier s statement that the rocess yields 3% defective roduct.

9 QReview TEST FOR GOODNESS OF FIT To test a samle distribution to determine if it fits the attern of a secific distribution such as the normal, a Goodness of Fit test is erformed. Goodness of Fit refers to the comarison of an observed samle distribution with a known theoretical distribution. In the following examle the chi square distribution is used to comare a samle distribution with the normal distribution. This is a test to determine if the universe from where the samle data were obtained could be normally distributed. Chi Square = χ = ( f F ) i i F f i = observed frequency F i = theoretical frequency Based on the samle data, a hyothesis that the universe may be normally distributed against an alternative that it is not normally distributed is to be tested. H 0 : universe may be normally distributed H : universe is not normally distributed For a chi square test involving intervals or cells, the degrees of freedom are k - 3, where k is the number of intervals. For tests using row and column matrices called contingency tables, the degrees of freedom are (# rows ) x (# columns ) or (r )(c ). Examle 4 The measurements from a samle of one hundred number 0 bearings from the Canae Bearing Comany are groued into seven intervals or cells. The data table is on the next age. The number of intervals is arbitrary. More cells will give a higher degree of accuracy. Six to Twelve cells will yield satisfactory results. The number of occurrences in each cell is comared to the number of occurrences that should be in the cell based on the known theoretical distribution. In this examle, the data are comared to the normal distribution. The chi square test sums the differences in each cell and the total value of chi square is used to test the hyothesis that the data may be normally distributed. The mean and standard deviation are calculated using the midoint (m i ) of the interval and the observed frequencies (f i ). fimi x = ( ) 30 = = 3. n 00 i fi mi x s = [ ( ) ] n 9339 = = Could the universe from where these data were obtained be normally distributed? Test at a level of significance of.05. The null and alternate hyotheses are H 0 : universe may be normally distributed H : universe is not normally distributed

10 Chater 5 Statistical Inference 79 Interval Observed Frequency (fi) Theoretical Frequency (Fi) ( fi Fi) F 40 or above < < < < < below χ =.4 The degrees of freedom are k - 3. There are seven intervals, therefore d.f. are 7-3 = 4 Chi square = χ =.4. From the chi square table, the critical value for α =.05 and 4 degrees of freedom is Therefore, the null hyothesis is not rejected. It is concluded that the universe could be normally distributed. 4. Sketch of the Chi Square Distribution The chi square distribution is not a symmetrical distribution. Its shae deends on the degrees of freedom. If it is known that a universe is normal, the distribution of samle variances has the form of the chi square distribution. The chi square values that are comuted in the examle are samle variances. This is why the chi square distribution is used to erform the test. The critical value of chi square is defined as the value on the abscissa that indicates where the region of rejection begins. The numbers on the abscissa are the various values of chi square. The region of rejection is the area to the right of the critical value and is indicated by the shading. i Region of Rejection α = χ If goodness of fit tests are made to comare samle data to other distributions, such as the uniform or binomial, the theoretical frequencies would be obtained using the secific distribution.

11 QReview Theoretical Frequencies for the Examle The theoretical frequencies are obtained from the standard normal curve table x Z x = 3. and s = INFERENCES ABOUT TWO MEANS When a control grou is comared with an exerimental grou, or when two exerimental grous are comared, the concern is rimarily with differences in their means. This test assumes that the means are indeendent. Two samles are indeendent when the outcome of one does not affect the other. The ooled variance is the combined variance of two or more samles and is denoted by s. The ooled variance for two samles is s = n ( n ) s + ( n ) s + n and n and s are the samle size and variance for the first samle and n and s are the samle size and variance for the second. The ooled standard deviation is s denominator (n + n - ) indicates the total degrees of freedom. = s. The

12 Chater 5 Statistical Inference 8 Two rocesses are being evaluated. A study shows that 0 emloyees working on rocess A have an average assembly time of 50 seconds with variance of 00. In addition, 0 emloyees on rocess B have an average assembly time of 5 seconds with variance of 80. Are the average assembly times for the two rocesses significantly different? For this tye of test, the variances are always tested first for significant differences. The F ratio and corresonding F table values are used to test the variances. If the variances are significantly different, the test for differences in means is not valid. Before testing for differences in means, the assumtion is that the variances are homogeneous. The t table is used for the test of differences between means since both samle sizes are less than thirty. The formula for conversion from the x scale to the t scale: t = s x x n + n The variances are tested for significant differences (α =.05). ) H o : σ = σ H : σ > σ sl 00 ) F = = =. 5 s 80 S The F ratio is a value on the abscissa of the F distribution curve. The symbol s L reresents the larger of the two variances and the symbol s s reresents the smaller of the two variances. The samle size for the larger variance is denoted by n L and the samle size for the smaller variance is denoted by n s. There are n L - degrees of freedom for the numerator and n s - degrees of freedom for the denominator. 3 ) Region of rejection: F.4. This is the critical value of F from the F.05 table with 9 degrees of freedom for the numerator and 9 degrees of freedom for the denominator. Region of Rejection α = F

13 QReview 8 4) The calculated value of F is less than the critical value of.4. The null hyothesis that σ = σ is not rejected. It is concluded that there is no significant difference between the two variances. The means are now tested for significant differences (α =.05). ) H o : µ = µ H : µ µ ) Region of rejection: > t > (These are the critical values of t). 3) s = n ( n ) s + ( n ) s + n 9( 00) + 9( 80), s = s = 8 = ) t = = = = s x x n + n (.3873) (Area = ½ α =.05) (Area = ½ α =.05) 0 5 d t 4) The calculated value of t, t = +6.94, falls beyond the critical value of t = +.048, therefore the null hyothesis that µ = µ is rejected. The conclusion is that there is a statistically significant difference between the two means.