Selecting the Appropriate Inferential Test:For each of the following: a) where applicable, identify the independent and dependent variables;b) write the null and the alternative hypotheses; c) find the critical value(s) and make sure to note whether thetest is one- or two-tailed; d) state what type of test to perform (e.g., one sample z, paired samples t test, OneWay ANOVA, etc.) Assume alpha = .05 for each problem.1) A study is conducted to determine how the stimulant drug Adderall impacts the ability to sustain attentionalvigilance. Twenty-two college students perform a computerized test of attentional vigilance two times; one testis performed immediately after taking Adderall and the other test is performed without the drug treatment. Tests1 and 2 are carried out one week apart. The researchers will test the hypothesis that Adderall improves vigilance. n(Ɛ D²)– (E D)²п -1n = number of pairs of scores (for example, if 8 people are measured at tl and t2, df = 8-1 = 7)D = the difference between each pair of scoresED = the sum of the differences(ED²) = the sum of the squares of the differences(ED)²= the sum of the differences squaredt score: Independent Samples t-test (degrees of freedom = (n1 – 1) + (n2 – 1))п, — 1-1= mean of first group (sample)X2= mean of second group (sample)s1 = sample variance of first groups?2 = sample variance of second groupn1 = number of subjects (or observations) in the first groupn2= number of subjects in second groupBe careful about computing the denominator; pay attention to whether you are working with sample variances or standarddeviations to make sure your computations are correct.One Way ANOVA (see ANOVA table for degrees of freedom)Between Groups Sum of Squares: Subtract the grand mean from each group mean, square the deviation,multiply by n in each group and add the values together.SSsΣη (X.-(X,n = the number of subjects or observations in a groupX, = the mean for any group (1 through n) of subjectsX g = the overall, or grand, mean for all subjects or observationsWithin Groups Sum of Squares: Subtract each group's mean from the individual scores and add them all up.SSWΣΣ (X- Χ.X = any single score within any particular groupX,= the mean of any particular groupnG = number of groupsN= total # of subjects/observationsdfMSFTreatmentF =MS3 =Ss, / dfgSS.dfg =ng-1(Between= B)MS /MSwError (Within= W) SSwdfw =N -ngMSw=SSw/ dfwTotalSSTotal =SSg + SSwdfrotal=N -1 =df,+ dfw

Given information: A study is conducted to determine how the stimulant drug Adderall impacts the…

Answered: Selecting the Appropriate Inferential…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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We have provided the numbers of successes and the sample sizes for independent simple random samples from two populations. In each case, do the following tasks. a. Determine the sample proportions. b. Decide whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d). c. Use the two-proportions z-test to conduct the required hypothesis test. d. Use the two-proportions z-interval procedure to find the specified confidence interval. x1 = 18, n1 = 40, x2 = 30, n2 = 40; left-tailed test, α = 0.10; 80% confidence interval
Find the P-value for the following values of the test statistic t, sample size n, and alternate hypothesis H. Use the Critical Values for the Student's t Distribution Table and specify that P is between two values. Part 1 of 4 When t -2.31, n 20, and H, Po 0.02 < P-value <0.05 Part: 1/4 Part 2 of 4 When t= -3.164, n 11, and H H < P-value <
Q.3 For the following data: a) Compute SST, SSTR, SSE b) Construct a one-way ANOVA table c) Decide at the 5% significance level, whether the data provide sufficient evidence to conclude that the means of the populations from which the samples are drawn are not the same. Sample 1 Sample 2 Sample 3 Sample 4 11 16 6 2 10 7 4 10 3. we have: ng = 3 n3= 3 n4 = 3 Ta 24 = 1s Ty= 36 THa9 n z Enj = 12 Exz= IT;= &4 ..... Ex - (Ex:*/n = SSTR = I (Tim;) - (Iri)/n = ()+( )+( )+( ) -- 2. SST = 12 SSE = 5sT - SSTR = Fostatistic df MS 2 df :عاطء Souree SS MSTR/MSE 138 Treatment where: MSE- SSE/(n-k) MSTR = SSTR/(K-1) Enron Tofal df z ( 3, 8) caitical Valme s ? Compare test- statistic to cv. Do you rejuctNull or Do not rejrit what dees it mean ? write here
The number of points held by random samples of the NHL's highest scorers of both the Eastern Conference and the Western Conference is shown. At the 0.05 level of significance, can it be concluded that there is a difference in means based on theses data: Eastern Conference Western Conference 83 60 75 58 77 59 72 58 78 59 70 58 37 57 66 55 62 61 59 61 What is the appropriate null hypothesis for this problem? (u= population mean for this problem) What is the critical Value? What is the test statistics? What is the decision?
Conduct the following test at the a= 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the critical value. Assume that the samples were obtained independer sampling. Test whether p, # p,. Sample data are x, = 30, n, = 255, x, = 38 and n, = 301. (a) Determine the null and alternative hypotheses. Choose the correct answer below. O Ho: p, = P2 versus H,: p, p2 O Ho: P, = P2 versus H,: p, # P2 (b) The test statistic z, is . (Round to two decimal places as needed.) (c) The critical values are +. (Round to three decimal places as needed.) Test the null hypothesis. Choose the correct conclusion below. O Reject the null hypothesis. O Do not reject the null hypothesis.
The Mann-Whitney U test is used to compare two independent samples. Given the following data, use the Mann-Whitney U test to determine if there is a significant difference between the two sets of scores: Sample 1: 2, 4, 5, 7, 8 Sample 2: 1, 3, 4, 5, 6 What is the U statistic for this data and is there a significant difference between the two samples?
Consider the following hypothesis test. Ho: H1 - H2 = 0 Ha: H1 - 42 # 0 The following results are from independent samples taken from two populations assuming the variances are unequal. Sample 1 Sample 2 n1 35 = 40 = 13.6 X, = 10.1 S1 5.8 S2 = 8.1 (a) What is the value of the test statistic? (Use x, - x2. Round your answer to three decimal places.) (b) What is the degrees of freedom for the t distribution? (Round your answer down to the nearest integer.) (c) What is the p-value? (Round your answer to four decimal places.) p-value = (d) At a = 0.05, what is your conclusion? Do not Reject Ho: There is sufficient evidence to conclude that u, - µ, ± 0. Reject Ho. There is sufficient evidence to conclude that µ, - µ, # 0. Do not Reject Ho. There is insufficient evidence to conclude that u, - H, + 0. 0' Reject Ho. There is insufficient evidence to conclude that u1 - µ2 ± 0.
Assume that a simple random sample has been selected from a normally distributed population and test the given claim.A researcher wants to test the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds that x = 21.3 months and s = 5.7 months. Test the claim at the 0.05 significance level.State the claim in words: Convicted Burglars spend an average of 18.7 months in jail. Ho μ = 18.7 H1 μ ≠ 18.7 What calculator function is needed? p value = __________ Since p ________ α, ________________________ the null hypothesis. Final conclusion:
The value obtained for the test statistic, z, in a one-mean z-test is given. Whether the test is two tailed, left tailed, or right tailed is also specified. For parts (a) and (b), determine the P-value and decide whether, at the 1% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. a. The test statistic in a two-tailed test is z = -0.93. The P-value is (Round to three decimal places as needed.) ... sufficient evidence to reject the null hypothesis in favor of the significance level. At the 1% significance level, the data the alternative hypothesis because the obtained P-value is b. The test statistic in a two-tailed test is z = 0.61. The P-value is (Round to three decimal places as needed.) sufficient evidence to reject the null hypothesis in favor of the significance level. At the 1% significance level, the data the alternative hypothesis because the obtained P-value is Time Remaining: 01:57:58 Mand
Provide an appropriate response. Test the hypothesis that p1 = P2. Use a = 0.05. The sample statistics listed below are from independent samples. Sample statistics: n1 = 50, x1 35, and n2 = 60, x2 = 40 Step 1) State the null and alternative hypotheses. Step 2) Determine the critical value for the level of significance, a. Step 3) Find the test statistic or P-value. Step 4) Will the researcher reject the null hypothesis or do not the null hypothesis: Step 5) Write the conclusion. HTML Editor BIUA A、IE三= Ex x E 曲。 12pt Pa SAMSUNG 10
The value obtained for the test statistic, z, in a one-mean z-test is given. Whether the test is two tailed, left tailed, or right tailed is also specified. For parts (a) and (b), determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. a. The test statistic in a right-tailed test is z=2.00. The P-value is . (Round to three decimal places as needed.)

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