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Differences in Means You use this tool if you want to know if one variable is typically bigger (smaller, etc.), ON AVERAGE, than another. You should choose the variable you think might be larger to be Variable 1. Look for KEY WORDS in the problem. It will ask whether one mean is greater than (>), less than (<), greater than or equal to (≥), less than or equal to (≤), the same as (=), or different from (≠) the other mean. You write what the question asks you to test EXACTLY, as if the statement was true. If the question Subtract the mean on the right from both sides, and simplify. (The right side of the equation will always have Mean2-Mean2 in it, which simplifies to 0.) To get the output, go to Data Analysis. Select t Test: Two Sample Assuming Unequal Variances or t-test Paired Two Sample for Means, depending on how the sample were selected. A researcher who the results are due to the one thing that is different between the two samples. For example, researchers might use identical twins to test the effects of a drug. These samples may be referred to Write H 0 and H 1 . There are only 3 possibilities: H 0 : Mean 1 -Mean 2 ≥ 0 H 1 : Mean 1 -Mean 2 < 0 or H 0 : Mean 1 -Mean 2 ≤ 0 H 1 : Mean 1 -Mean 2 > 0 or H 0: Mean 1 -Mean 2 = 0 H 1 : Mean 1 -Mean 2 ≠ 0 asks, "Is Mean 1 greater than Mean 2 ", you would write Mean 1 > Mean 2 Example: Mean 1 -Mean 2 > Mean 2 -Mean 2 simplifies to Mean 1 -Mean 2 > 0 Now write the complement to the original statement. The complement must include ALL possible states of the world not included in the first statement. Example: If Mean1-Mean2 is not > 0, it must be ≤ 0 You would write the second hypothesis as Mean 1 -Mean 2 ≤ 0 NOW, and ONLY NOW, decide which is H 0 . It will be the hypothesis with the equal sign in it. Label the other hypothesis H 1 . H 0 : Mean 1 -Mean 2 ≤ 0 H 1 : Mean 1 -Mean 2 > 0 uses the scientific method will keep all RELEVANT factors that might affect the results SUFFICIENTLY the same, except for one variable that is different. That way, he or she knows that as matched, paired or dependent samples. For example, the researcher may select a random sample of

items, and then select a second sample that has the same relevant characteristics as the first sample. random samples are sometimes referred to as independent samples. For the Variable 1 range, select all the items from the variable that you suspect will have the largest mean. Select all the items for the second variable for Variable 2 range. The hypothesized difference is the number on the right of the equation you have for H0. The alpha is the level of significance. Check labels if you selected the label as part of your input range. Otherwise, don’t. below: than the pvalue you wrote for the last step. If it is not greater than the pvalue, then you do not reject "Yes" or "No" will ALWAYS BE MARKED WRONG! Your choices are "Reject" or "Not reject." You should always test before you say whether two means are equal or different from each other. If the sample is collected this way, use t-test: Paired Two Sample for Means. If the researcher selects TWO random samples, you should use t test: Two Sample Assuming Unequal Variances. Two Write the pvalue. You use P(T<=t) one-tail if your H 1 had > or < in it, like the ones H 0 : Mean 1 -Mean 2 ≥ 0 H 1 : Mean 1 -Mean 2 < 0 or H 0 : Mean 1 -Mean 2 ≤ 0 H 1 : Mean 1 -Mean 2 > 0 You use P(T<=t) two-tail if your H 0 had = in it, like the H 0 below: H 0 : Mean 1 -Mean 2 = 0 H 1 : Mean 1 -Mean 2 ≠ 0 The pvalue tells you how likely it was to get the results that you actually got IF H 0 is true. If it is very likely that you would get the results you got, you would be willing to think that H 0 was the true statement. If it is very unlikely to get the results you got if H 0 was true, you would be willing to think that H 0 was false, and that H 1 was the true statement. How low does the probability have to be for you to be willing to reject H 0 in favor of H 1 ? The level of significance is the standard. If the pvalue is less than the level of significance, we reject H 0 . Write whether you rejected or did not reject H 0 . You reject H 0 if the level of significance is greater H 0 . Now that you know which statement you will treat as being true (H 0 or H 1 ), answer the question. Answer the question based on which statement you treat as true. If you reject H 0 , act as if H 1 is true. If you do not reject H 0 , act as if H 0 is true.

Practice Problems 1.The table below shows SAT scores before and after the sample of 10 students took a preparatory course. Is there sufficient evidence to conclude that the preparatory course is effective in raising scores? Use a .05 significance level. Student SAT Before Course SAT After Course A 700 720 B 840 840 C 830 820 D 860 900 E 840 870 F 690 700 G 830 800 H 1180 1200 I 930 950 J 1070 1080 t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 888 877 Variance 24084.4444444444 22845.5555555556 Observations 10 10 Pearson Correlation 0.991609165037859 Hypothesized Mean Difference 0 df 9 t Stat 1.71791138077467 P(T<=t) one-tail 0.059968753602172 t Critical one-tail 1.83311293265624 (a) Write H 0 and H 1 . (1 pt) (b) Get the output. (1 pt)

P(T<=t) two-tail 0.119937507204344 t Critical two-tail 2.26215716279821 (e) There (is / is not) sufficient evidence to conclude that the (f) Would your answer to part (d) and (e) change if the level of 2. A manufacturer is developing a nickel-metal hydride battery to be used in cellular phones. The director of quality control wants to determine whether the new battery is better than the old nickel-cadmium battery. The performance measure of interest is the talking time (in minutes) before recharging. Nickel-Cadmium Battery Nickel -Metal Hydride Battery 54.5 78.3 67.8 95.4 64.5 69.4 70.4 87.3 72.5 62.5 64.9 85.0 83.3 85.3 72.8 72.1 68.8 41.1 71.0 103.0 41.7 81.3 69.7 46.4 40.8 82.3 (c) State the pvalue. (1 pt) (d) State whether you reject or do not reject H 0 . (1 pt) preparatory course is effective in raising scores. (1 pt) significance was .10? Yes No (1 pt)

(e) At the .05 level of significance, the nickel-metal battery (is / is not) Answers (b) t-Test: Paired Two Sample for Means SAT Before Course SAT After Course Mean 877 888 Variance 22845.5555555556 24084.4444444444 Observations 10 10 Pearson Correlation 0.991609165037859 Hypothesized Mean Difference 0 df 9 t Stat -1.71791138077467 P(T<=t) one-tail 0.059968753310091 t Critical one-tail 1.83311385626439 P(T<=t) two-tail 0.119937506620183 t Critical two-tail 2.26215888687875 (c) pvalue = .059968753 (1 pt) (1 pt) (e) There's no evidence that preparatory course is effective in raising scores. (f) Yes (1 pt) (c) State the pvalue. (1 pt) (d) State whether you reject or do not reject H 0 . (1 pt) better than the nickel-cadmium battery. (1 pt) (1) (a) H 0 : MeanBefore - Mean After > 0 or H 0 : MeanAfter - MeanBefore < 0 H 1 : MeanBefore - MeanAfter < 0 H 1 : MeanAfter - MeanBefore > 0 (d) Do not reject H 0

2. (a) MeanNickel-Cadmium < MeanNickel-Metal MeanNickel-Cadmium - MeanNickel-Metal < 0 (b) t-Test: Two-Sample Assuming Unequal Variances Nickel-Cadmium Battery Nickel -Metal Hydride Battery Mean 70.748 79.728 Variance 195.788433333333 226.029600000002 Observations 25 25 Hypothesized Mean Difference 0 df 48 t Stat -2.18616919155522 P(T<=t) one-tail 0.016855863047736 t Critical one-tail 1.67722419064376 P(T<=t) two-tail 0.033711726095471 t Critical two-tail 2.01063357962994 H 0 :MeanNickel-Cadmium - MeanNickel-Metal > 0 or H 0 : m NickelMetal - m NickelCadmium < 0 H 1 :MeanNickel-Cadmium - MeanNickel-Metal < 0 H 1 : m NickelMetal - m NickelCadmium > 0 (c) p value = .01685586 (1 pt) (d) Reject H0. (1 pt) (e) The nickel-metal is better. (1 pt)

times of employees trained in a computer-assisted individual-based program Yes No 2. A study was conducted to investigate some effects of physical training. The weights of the subjects before and after training are given below (all weights are given in kilograms). Is there sufficient evidence to conclude that there is a difference between the pretraining and posttraining weights? Test at the .05 level of significance. Subject PreTraining PostTraining A 90 92 B 57 55 C 62 58 D 69 69 E 74 66 F 77 76 G 59 58 H 92 88 I 70 69 J 85 84 (1 pt) (d) At the .05 level of significance, there (is / is not) evidence that the average assembly is less than the assembly times for those trained in a team based program. (1 pt) (e) Is the computer-based training better than the team-based training? (1 pt) (a) Write H 0 and H 1 . (b) Get the output. State the pvalue. (1 pt)

(d) There (is / is not) a difference between the pretraining and posttraining weights Answers t-Test: Two-Sample Assuming Unequal Variances Computer-Assisted Team-Based Mean 17.5571428571429 18.48571428571 Variance 3.73757142857135 17.52728571429 Observations 21 21 Hypothesized Mean Differenc 0 df 28 t Stat -0.922770557633667 P(T<=t) one-tail 0.182005281054105 t Critical one-tail 1.70113025887986 P(T<=t) two-tail 0.36401056210821 t Critical two-tail 2.04840944206808 pvalue = 0.182005281054105 (c) State whether you reject or do not reject H 0 . (1 pt) at the .05 level of significance. (1 pt) 1. (a) State H 0 and H 1 (1 pt) H 0 : Mean Computer -Mean Team > 0 or H 0 : Mean Team -Mean Computer < 0 H 1 : Mean Computer -Mean Team < 0 H 1 : Mean Team -Mean Computer > 0 (b) Get the output. State the pvalue. (1 pt) (c) State whether you reject or do not reject H 0 (1 pt). Do not reject H 0

Team-Based 21.4 18.7 19.3 15.6 16.1 21.7 30.7 13.8 18 14.8 17.1 18.2 18.6 24.7 15.3 17.4 23.2 20.1 12.3 15.2 16

Reject H 0

is not 2. Is there a difference between "ave respondents"? A well-known perso both average Americans and phon personality factor, conscientiousne conscientiousness. At the .05 leve have higher conscientiouness scor conscientiousness scores between Phone Survey Respondents Average Americans a. 1 2 3 4 5 b. t-Test: Two-Sample Assuming Unequal Varian Phone Survey Respondent Mean 37.473 Variance 3.85775666666667 Observations 10 Hypothesized Mean D 0 df 16 t Stat 2.35099979319465 P(T<=t) one-tail 0.015936312872761 t Critical one-tail 1.74588367627625 P(T<=t) two-tail 0.031872625745522 t Critical two-tail 2.11990529922125 c. d. e. At the .05 level of significance, the are 3. Ten bank tellers are asked to help speed and accuracy with which th They are first given a five-minute State H 0 and H 1 . Your MUST show Mean Phone Survey Respondents > Mean Mean Phone Survey Respondents - Mean A Mean Phone Survey Respondents - Mean A Mean Phone Survey Respondents - Mean A H 0 : Mean Phone Survey Respondents - Me H 1 : Mean Phone Survey Respondents - Me H 0 : Mean Phone Survey Respondents - Me H 1 : Mean Phone Survey Respondents - Me Get the Excel Output (.5 pt.) What is your pvalue? (1 pt.) Did you reject or not reject H 0 ? (1 (are / are not) higher, on average,

Drill" test. The corrective drill is ex substantial improvement in perfor At the .05 level of significance, did more than 2 points? (Higher score Be careful is setting up H0 and hypothesized mean difference Teller Before Drill Score A 45 B 52 C 34 D 38 E 47 F 42 G 61 H 53 I 52 J 49 a. 1 2 3 4 5 b. t-Test: Paired Two Sample for Means After Drill Score Mean 50.8 Variance 83.7333333333332 Observations 10 Pearson Correlation 0.968709651135698 Hypothesized Mean D 2 df 9 t Stat 1.92758821595154 P(T<=t) one-tail 0.043002183938058 t Critical one-tail 1.83311293265624 P(T<=t) two-tail 0.086004367876115 t Critical two-tail 2.26215716279821 State H 0 and H 1 . Your MUST show Mean After Drill > Mean Before Drill Mean After Drill - Mean Before Drill > 2 Mean After Drill - Mean Before Drill > 2 Mean After Drill - Mean Before Drill ≤ 2 H 0 : Mean After Drill - Mean Before Drill ≤ H 1 : Mean After Drill - Mean Before Drill > H 0 : Mean After Drill - Mean Before Drill ≤ H 1 : Mean After Drill - Mean Before Drill > Get the Excel Output (.5 pt.)

Homework 7 SP23 2nd 16 pts. about the length of time a particular drug retains its ottles of the product is drawn from current production nd sample is obtained from the same batch, stored for readings obtained are as follows. Higher readings B C D E F G H I 10.7 10.3 10.2 9.8 10.6 10.4 9.8 10.2 10.5 10.1 9.8 9.6 10.5 10.3 9.6 10.3 Stored Samples 10.088888888889 0.1236111111111 9 0.008079082 ean potency the same after one year? Test at the .05 the five-step process for determining H 0 and H 1 (1 pt.) ples es ≠ 0 es ≠ 0 es = 0 ed samples = 0 d samples ≠ 0 ed samples = 0 d samples ≠ 0 Reject H 0 he same after one year. (1 pt) . ≠

erage Americans" and those who are "phone survey onality survey is designed to assess the personality profile of ne survey rspondents. The results of the survey for one ess, are displayed below. A higher score indicates more el of significance, test whether "phone survey respondents" res, on average, than "average Americans." n "average Americans" and "phone survey respondents. 37.38 35.06 35.74 38.97 37.84 36.5 41.7 36.31 36.93 36.9 33.56 37.24 34.59 34.95 37.46 34.73 nces Average Americans 35.698 1.8424622222222 10 0.015936313 e conscientiousness scores of "phone survey respondents" p test whether a corrective drill significantly improves the hey enter customer transactions into the bank computer. "Before Drill" test, then the drill, and finally an "After the five-step process for determining H 0 and H 1 (1 pt.) Average Americans Average Americans > 0 Average Americans > 0 Average Americans ≤ 0 ean Average Americans ≤ 0 ean Average Americans > 0 ean Average Americans ≤ 0 ean Average Americans > 0 Reject H 0 , than the conscientiousness scores of average Americans. (1 pt.)